HW 3 - Linear Maps

3.A: The Vector Space of Linear Maps

#1

Theorem

Suppose $b, c \in R$ . Define $T : R^{3} \to R^{2}$ by:

T (x, y, z) = (2 x - 4 y + 3 z + b, 6 x + c x y z)

Then $T$ is linear iff $b = c = 0$

Proof
Consider $\to$ first. Suppose that $T$ is linear. Then any $T ((x_{1}, y_{1}, z_{1}) + (x_{2}, y_{2}, z_{2})) = T (x_{1}, y_{1}, z_{1}) + T (x_{2}, y_{2}, z_{2})$ and similar for homogeneity.

Notice that:

T (0, 0, 0) = (b, 0) = \vec{0}

since all $T (0) = 0$ via 3.11, so then we require that $b = 0$ . Furthermore, notice that:

T (1, 1, 1) = (2 - 4 + 3 + b, 6 + c)

but since $b = 0$ then:

T (1, 1, 1) = (1, 6 + c)

And notice that:

T (1, 1, 1) = T ((1, 1, 0) + (0, 0, 1)) = T (1, 1, 0) + T (0, 0, 1) = (- 2, 6) + (3, 0) = (1, 6)

so equating parts of our vector then $6 + c = 6 \Rightarrow c = 0$ .

Now consider $\leftarrow$ , so suppose that $b = c = 0$ , then:

T (x, y, z) = (2 x - 4 y + 3 z, 6 x)

Let $u, v \in R^{3}$ be arbitrary. Notice that:

\begin{aligned} T (u + v) & = T ((u_{x} + v_{x}, u_{y} + v_{y}, u_{z} + v_{z})) \\ = (2 (u_{x} + v_{x}) - 4 (u_{y} + v_{y}) + 3 (u_{z} + v_{z}), 6 (u_{x} + v_{x})) \\ = ([2 u_{x} - 4 u_{y} + 3 u_{z}] + [2 v_{x} - 4 v_{y} + 3 v_{z}], 6 (u_{x}) + 6 (v_{x})) \\ = (2 u_{x} - 4 u_{y} + 3 u_{z}, 6 u_{x}) + (2 v_{x} - 4 v_{y} + 3 v_{z}, 6 v_{x}) \\ = T u + T v \end{aligned}

so we have additivity. Let $α \in R$ be arbitrary. Then:

\begin{aligned} α T (u) & = α (2 u_{x} - 4 u_{y} + 3 u_{z}, 6 u_{x}) \\ = (2 α u_{x} - 4 α u_{y} + 3 α u_{z}, 6 α u_{x}) \\ = (2 (α u_{x}) - 4 (α u_{y}) + 3 (α u_{z}), 6 (α u_{x})) \\ = T (α (u_{x}, u_{y}, u_{z})) \\ = T (α u) \end{aligned}

showing homogeneity, thus $T$ is a linear map, which completes the proof.
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#4

Theorem

Suppose $T \in L (V, W)$ and $v_{1}, . . ., v_{m}$ is a list of vectors in $V$ such that $T v_{1}, . . ., T v_{m}$ is a LI list in $W$ . Then $v_{1}, . . ., v_{m}$ is LI.

Proof
Consider the sum:

a_{1} v_{1} + \dots + a_{m} v_{m} = \vec{0}

Notice that we can take $T$ on both sides:

T (a_{1} v_{1} + \dots + a_{m} v_{m}) = T (\vec{0}) = \vec{0}

but by the linearity of $T$ :

T (a_{1} v_{1} + \dots + a_{m} v_{m}) = a_{1} (T v_{1}) + \dots + a_{m} (T v_{m}) = \vec{0}

But we know that $T v_{1}, . . ., T v_{m}$ is a LI list in $W$ , so then the only constant choice for $a_{i}$ is all $a_{i} = 0$ , so then our initial list $v_{1}, . . ., v_{m}$ is LI by proxy.
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#7

Theorem

Consider $V$ where $dim (V) = 1$ , and $T \in L (V, V)$ . Then there exists $λ \in F$ such that $T v = λ v$ for all $v \in V$ .

Proof
Since $dim (V) = 1$ then there is some basis $v$ such that $v \in V$ , so $V = span (v)$ and $v \neq \vec{0}$ . Let $x \in V$ be arbitrary, so then:

x = α v

since $x \in span (v)$ . Now, compose $T$ on both sides:

T (x) = T (α v) = α (T v)

but since $T v \in V$ then we can just see that $x = α v = T v$ so then we can choose $λ = α$ and we are done.
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Proof
We just need to show that either additivity or homogeneity (one of them) are, for some vectors, not satisfied. There's some vector $u \in U$ such that, since $S \neq 0$ then $S u \neq 0$ . Also since $U \neq V$ then there is some vector $v \in V$ while $v \notin U$ .

Consider $T v$ , since $v \in V$ . Notice that $v \in V$ while $v \notin U$ so then $T v = {\vec{0}}_{W}$ . Further, since $U$ is a subspace of $V$ then $u \in V$ , so consider $T u$ . Since $u \in U$ then $T u = S u \neq {\vec{0}}_{W}$ . Consider:

T (u + v)

Since $v \notin U$ then since $v$ is created based off of $V$ 's basis, while not being spanned by $U$ 's basis, then $u + v \notin U$ . Clearly $u + v \in V$ though so then $T (u + v) = {\vec{0}}_{W}$ by definition. But notice that, if we assume for contradiction that additivity holds under $T$ , then:

T (u + v) = T u + T v = w + {\vec{0}}_{W} = w

where since $T u \neq {\vec{0}}_{W}$ then $w$ is some non-zero vector. But look! We had it that $T (u + v) = {\vec{0}}_{W} = w$ but $w$ is non-zero. Hence, our assumption was false. We found a counterexample to $T$ having additivity, and thus $T$ isn't a linear map on $V$ .
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#11

Theorem

Suppose $V$ is finite-dimensional, so $dim (V) = n$ . Then any $T \in L (U, U)$ for any subspace $U$ of $V$ can be extended to a linear map on $V$ . In other words, if $U$ is a subspace of $V$ and $S \in L (U, W)$ then there is some $T \in L (V, W)$ such that $T u = S u$ for all $u \in U$ .

Proof
Let $v_{1}, . . ., v_{k}$ be a basis for $U$ whos dimension is $k \leq n$ . We can always extend it to a basis $v_{1}, . . ., v_{k}, v_{k + 1}, . . ., v_{n}$ which is a basis for $V$ . Define:

w_{1} = S (v_{1}), . . ., w_{k} = S (v_{k})

then define $T$ such that:

T (v_{1}) = w_{1}, . . ., T (v_{k}) = w_{k}, T (v_{k + 1}) = \vec{0}, . . ., T (v_{n}) = \vec{0}

Let $u \in U$ be arbitrary. Then:

\begin{aligned} T (u) & = T (a_{1} v_{1} + \dots + a_{k} v_{k}) \\ = a_{1} T v_{1} + \dots + a_{k} T v_{k} \\ = a_{1} w_{1} + \dots + a_{k} w_{k} \\ = a_{1} S (v_{1}) + \dots + a_{k} S (v_{k}) \\ = S (a_{1} v_{1} + \dots + a_{k} v_{k}) \\ = S (u) \end{aligned}

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3.B: Null Spaces and Ranges

#1

A linear maps $T$ where $dim (null (T)) = 3$ and $dim (range (T)) = 2$ is the map $T : R^{5} \to R^{2}$ where:

T (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = T (x_{1}, x_{2})

Because $null (T) = {v \in V : T v = 0} = span ((0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 0, 0, 0, 1))$ and thus our dimension there is correct. Here also, we can just use the fundamental theory of linear maps to know our other value for the dimension of range is correct.

#2

Theorem

$V$ is a vector space and $S, T \in L (V, V)$ where $range (S) \subset null (T)$ . Then $(S T)^{2} = 0$ , the zero map.

Proof
We need to prove that for some arbitrary vector $v \in V$ that $(S T)^{2} (v) = 0$ . Since $range (S) \subset null (T)$ then for any vector like $v \in V$ then $T S v = 0$ because $S v \in range (S)$ implies that $S v \in null (T)$ so then $T (S v) = 0$ .

As such:

(S T)^{2} (v) = S (T S) T v = S (T S v^{'}) = S (0) = 0

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#3

Suppose that $v_{1}, . . ., v_{m}$ is a list of vectors in $V$ , and $T \in L (F^{m}, V)$ by:

T (z_{1}, . . ., z_{m}) = z_{1} v_{1} + \dots + z_{m} v_{m}

a

The idea that $v_{1}, . . ., v_{m}$ spans $V$ comes from the surjectivity of $T$ as iff $v_{1}, . . ., v_{m}$ span $V$ then:

T (z_{1}, . . ., z_{m}) = V = z_{1} v_{1} + \dots + z_{m} v_{m}

So for the vector $\vec{z} = (z_{1}, . . ., z_{m})$ then $T \vec{z} \in V = span (v_{1}, . . ., v_{m})$ so then always for $w = T \vec{z}$ on our range we can always choose a vector $\vec{z}$ such that $T \vec{z} = \vec{w}$ showing surjectivity.

b

The idea that $v_{1}, . . ., v_{m}$ is LI on $V$ comes from the injectivity of $T$ . If $T$ is injective then given that $T ({\vec{z}}_{1}) = T ({\vec{z}}_{2}) = \vec{0}$ then that implies that ${\vec{z}}_{1} = {\vec{z}}_{2} = \vec{0}$ so then all $z_{1}, . . ., z_{m}$ for both are 0's, showing LI for the $v_{1}, . . ., v_{m}$ list. Going the other way, if $v_{1}, . . ., v_{m}$ is LI then the only way where $T (\vec{z}) = \vec{0}$ would be $\vec{z} = \vec{0}$ , implying that if $T ({\vec{z}}_{1}) = T ({\vec{z}}_{2}) = \vec{0}$ then ${\vec{z}}_{1} = \vec{0} = {\vec{z}}_{2}$ implying LI.

#9

Theorem

Suppose $T \in L (V, W)$ is injective and $v_{1}, . . ., v_{n}$ is LI in $V$ . Then $T v_{1}, . . ., T v_{n}$ is LI in $W$ .

Proof
Consider when:

a_{1} T v_{1} + \dots + a_{n} T v_{n} = \vec{0} \in W

Since $T$ is a linear map then:

T (a_{1} v_{1} + \dots + a_{n} v_{n}) = \vec{0}

implying that since $T$ is injective, then $null (T) = {\vec{0}}$ so then $a_{1} v_{1} + \dots + a_{n} v_{n} = \vec{0}$ . Since $v_{1}, . . ., v_{n}$ is LI then all $a_{i} = 0$ .
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#10

Theorem

Suppose $v_{1}, . . ., v_{n}$ spans $V$ and $T \in L (V, W)$ . Then $T v_{1}, . . ., T v_{n}$ spans $range (T)$ .

Proof
Consider:

v = b_{1} T v_{1} + \dots + b_{n} T v_{n}

by $T$ being linear:

v = T (b_{1} v_{1} + \dots + b_{n} v_{n})

so since $v_{1}, . . ., v_{n}$ spans $V$ then the vector $u = b_{1} v_{1} + \dots + b_{n} v_{n} \in V$ so then by the definition of what range is and $u \in V$ then $T u = v \in range (V)$ so since $v$ was an arbitrary vector from the span of $T v_{1}, . . ., T v_{n}$ then these vectors span the range of $T$ .
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#13

Theorem

Suppose $T$ is a linear map from $F^{4}$ to $F^{2}$ where:

null (T) = {(x_{1}, x_{2}, x_{3}, x_{4}) \in F^{4} : x_{1} = 5 x_{2}, x_{3} = 7 x_{4}}

Then $T$ is surjective.

Proof
We'll show that $T$ is surjective by showing that $range (T) = F^{2}$ .

Notice that:

\begin{aligned} null (T) & = {(x_{1}, x_{2}, x_{3}, x_{4}) \in F^{4} : x_{1} = 5 x_{2}, x_{3} = 7 x_{4}} \\ = span ((5, 1, 0, 0), (0, 0, 7, 1)} \end{aligned}

Notice that $dim (F^{4}) = 4$ and $dim (null (T)) = 2$ since the basis for $null (T)$ has 2 vectors, so then $dim (range (T)) = 2$ by the FTOLM. But notice that since $range (T) \subseteq F^{2}$ while being two dimensional then we must have $range = F^{2}$ showing surjectivity.
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#14

Theorem

Suppose $U$ is a 3-dimensional subspace of $R^{8}$ and that $T$ is a linear map from $R^{8} \to R^{5}$ such that $null (T) = U$ . Then $T$ is surjective.

Proof
Let $w \in R^{5}$ be arbitrary. We'll need to find some $v \in R^{8}$ such that $T w = v$ .

Notice that since $U$ is 3-dimensional then $dim (null (T)) = 3$ and by the FTOLM then $dim (range (T)) = 5$ as $dim (R^{8}) = 8$ . As such, $range (T)$ has some basis of 5 vectors:

w_{1}, w_{2}, w_{3}, w_{4}, w_{5}

We know that this basis has 5 vectors and this is a basis so there's LI of these vectors in $range (T)$ , but then since $dim (R^{5}) = 5$ then this is actually a basis for $R^{5}$ , so then $range (T) = R^{5}$ and thus $T$ is surjective.
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#15

Theorem

There doesn't exist a linear map from $F^{5}$ to $F^{2}$ whose null space equals:

{(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) \in F^{5} : x_{1} = 3 x_{2} \land x_{3} = x_{4} = x_{5}}

Proof
Assume for contradiction that such a linear map exists. Clearly we know $dim (F^{5}) = 5$ and notice that then:

dim (null (T)) = 2

as a valid basis for $null (T)$ is ${(1, 3, 0, 0, 0), (0, 0, 1, 1, 1)}$ , which we gloss over here. But notice that:

dim (F^{5}) = dim (range (T)) + dim (null (T))

thus plugging in our values we get that $dim (range (T)) = 3$ . But notice that if a linear map existed it would need a dimension of $dim (F^{2}) = 2$ instead! This is a contradiction, so $T$ cannot exist.
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#17

Theorem

Suppose $V$ and $W$ are both finite-dimensional. Then there exists an injective linear map from $V$ to $W$ iff $dim (V) \leq dim (W)$ .

Proof
Consider $\to$ , so suppose there's some $T$ injective linear map from $V$ to $W$ . By the FTOLM then:

dim (V) = dim (null (V)) + dim (range(V))

but since $T$ is injective, then $dim (T) = 0$ as $null (V) = 0$ so then since $dim (W) \geq dim (range (V))$ as $range (T)$ is a subspace of $W$ , then:

dim (V) = dim (range (V)) \leq dim (W)

Now consider $\leftarrow$ , so suppose that $dim (V) \leq dim (W)$ . Then there exist bases $v_{1}, . . ., v_{n}$ and $w_{1}, . . ., w_{m}$ for $n \leq m$ . Define the linear transformation $T : V \to W$ such that:

T (v_{1}) = w_{1}, \dots, T (v_{n}) = w_{n}

Now consider when $T (v) = T (u)$ where $v, u \in V$ are arbitrary. Since $v, u \in V$ then we can write them as their basis vectors:

T (v) = T (a_{1} v_{1} + \dots + a_{n} v_{n}) = a_{1} T v_{1} + \dots + a_{n} T v_{n}

and likewise for $u$ :

T (u) = T (b_{1} v_{1} + \dots + b_{n} v_{n}) = b_{1} T v_{1} + \dots + b_{n} T v_{n}

But since $T v_{j} = w_{j}$ then both:

\begin{aligned} T (v) & = T (u) \\ a_{1} T v_{1} + \dots + a_{n} T v_{n} & = b_{1} T v_{1} + \dots + b_{n} T v_{n} \\ a_{1} w_{1} + \dots + a_{n} w_{n} & = b_{1} w_{1} + \dots + b_{n} w_{n} \\ (a_{1} - b_{1}) w_{1} + \dots + (a_{n} - b_{n}) w_{n} & = \vec{0} \end{aligned}

but since $w_{1}, . . ., w_{n}$ is a basis then all $a_{i} - b_{i} = 0$ so $a_{i} = b_{i}$ so then $u = v$ showing $T$ as injective.
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#18

Theorem

Suppose $V$ and $W$ are both finite-dimensional. Then there is a surjective linear map from $V$ to $W$ iff $dim (V) \geq dim (W)$

Proof
Consider $\to$ , so $T : V \to W$ is surjective. Then by the FTOLM then:

dim (V) = dim (null (T)) + dim (range (T))

Since $T$ is a surjective map then $range (T) = W$ so then their dimensions are the same. Hence:

dim (V) = dim (null (T)) + dim (W) \geq dim (W)

Now consider $\leftarrow$ , so $dim (V) \geq dim (W)$ . Then $v_{1}, . . ., v_{n}$ is a basis for $V$ and $w_{1}, . . ., w_{m}$ is a basis for $W$ where $n \geq m$ . Define $T : V \to W$ as:

T v_{1} = w_{1}, . . ., T v_{m} = w_{m}, T v_{m + 1} = \vec{0}, . . ., T v_{n} = \vec{0}

Now consider some $w \in W$ . We will need to find some $v \in V$ such that $T v = w$ . Since $w \in W$ then we can write it as its basis vectors:

w = a_{1} w_{1} + \dots + a_{m} w_{m} + a_{m + 1} \vec{0} + \dots + a_{n} \vec{0}

it may seem arbitrary why I add the $\vec{0}$ on the right, but notice we can make a correspondance to each $w_{i}$ to $T v_{i}$ :

w = a_{1} T v_{1} + \dots + a_{m} T v_{m} + a_{m + 1} T v_{m + 1} + \dots + a_{n} T v_{n}

and since $T$ itself is a linear transformation:

w = T (a_{1} v_{1} + \dots + a_{m} v_{m} + a_{m + 1} v_{m + 1} + \dots + a_{n} v_{n})

And thus we can choose $v = a_{1} v_{1} + \dots + a_{m} v_{m} + a_{m + 1} v_{m + 1} + \dots + a_{n} v_{n} \in V$ as it's covered by the basis. Hence $T$ is surjective.
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3.C: Matrices

#1

Theorem

Suppose $V, W$ are finite-dimensional and $T \in L (V, W)$ . Then with respect to each choice of bases of $V$ and $W$ , the matrix of $T$ has at least $dim (range (T))$ nonzero entries.

Proof
Since $V, W$ are finite-dimensional then $v_{1}, . . ., v_{n}$ is a basis for $V$ and $w_{1}, . . ., w_{m}$ is a basis for $W$ where $dim (V) = n, dim (W) = m$ . Denote $B_{v}$ and $B_{w}$ as these bases. Furthermore, since $range (T)$ is a subspace of $W$ then $dim (range (T)) \leq dim (W) = m$ . Notice that the matrix for $T$ must be:

M (T, B_{v}, B_{w})

must be at least $n m$ entries as those are the sizes of the dimensions of $V$ and $W$ respectively. Notice though that $n m \geq m$ as $n \geq 1$ so then:

n m \geq m \geq dim (range (T))

where the LHS represents the number of entries in our matrix, so:

# Entries \geq dim (range (T))

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#2

Theorem

Suppose $D \in L (P_{3} (R), P_{2} (R))$ where $D p = p^{'}$ . We'll find a basis for our domain and co-domain such that the matrix of $D$ w.r.t. these bases is:

[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix}]

Proof
Let's construct such bases shall we? We need to have 4 basis vectors for our domain, and 3 for our codomain, labeled $p_{1}, . . ., p_{4}$ for the former and $q_{1}, . . ., q_{3}$ for the latter. Notice that then:

D (p_{1}) = q_{1}, D (p_{2}) = q_{2}, D (p_{3}) = q_{3}

while $p_{4}$ can be anything. Let's have $p_{1}, . . ., p_{4}$ be the standard basis for $P_{3} (R)$ , so then:

p_{1} = x, p_{2} = x^{2}, p_{3} = x^{3}, p_{4} = 1

Then notice that:

D (p_{1}) = q_{1} = 1, D (p_{2}) = q_{2} = 2 x, D (p_{3}) = q_{3} = 3 x^{2}

while $p_{4} = 1$ . Notice that since $D (p_{i}) = q_{i}$ we've by definition constructed a valid matrix that we needed.

We'd just need to show that ${1, 2 x, 3 x^{2}}$ is a valid basis for our codomain. Notice we have the right number of vectors via our dimension, so then we can just show LI:

a_{1} (1) + a_{2} (2 x) + a_{3} (3 x^{2}) = 0 + 0 x + 0 x^{2} \to a_{i} = 0

so it's LI and thus a valid basis.
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#3

Theorem

Suppose $V, W$ are finite-dimensional and $T \in L (V, W)$ . Then there exists a basis of $V$ and one for $W$ such that w.r.t these bases, all entries of $M (T)$ are 0 except that the entries in row $j$ and column $j$ , equal 1 for $1 \leq j \leq dim (range (T))$ .

Proof
Say $V$ is $n$ dimensional and $W$ is $m$ dimensional, so our matrix $M$ is $m \times n$ . Say that $k = dim (range (T)) \leq dim (W) = m$ . To get just ones on the diagonal for $M$ , we need to choose bases such that:

T (v_{1}) = w_{1}, . . ., T (v_{k}) = w_{k}, T (v_{k + 1}) = \vec{0}, . . ., T (v_{n}) = \vec{0}

where all the $v_{i}, w_{j}$ compose their respective bases. As such, start with a basis $w_{1}, . . ., w_{k}$ for $range (T)$ , which we can extend to a basis for $W$ as $w_{1}, . . ., w_{k}, . . ., w_{m}$ since $k \leq m$ . Then notice that there's some basis for this range $v_{1}, . . ., v_{k}$ . We can extend this basis to $v_{1}, . . ., v_{k}, . . ., v_{n}$ as a basis for $V$ . As such, then:

T (v_{1}) = w_{1}, . . ., T (v_{k}) = w_{k}

by definition of our how we chose our bases, as well as:

T (v_{k + 1}) = \vec{0}, . . ., T (v_{n}) = \vec{0}

since our extended vectors must be in the null space (they're not in the range as their intersection is the set with the zero vector). Notice that $v_{k + 1}, . . ., v_{n}$ is a basis for $null (T)$ .
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#4

Theorem

Suppose $v_{1}, . . ., v_{m}$ is a basis of $V$ , and $W$ is finite dimensional. Suppose $T \in L (V, W)$ . Then there is some basis $w_{1}, . ., w_{n}$ of $W$ such that all the entries in the first column of $M (T)$ w.r.t our bases are 0 except for possibly a 1 in the first row, first column.

Proof
We need to show some basis for $W$ such that:

T (v_{1}) = d w_{1}, T (v_{2}) = \vec{0}, . . ., T (v_{m}) = \vec{0}

where $d \in {0, 1}$ . We are given the basis $v_{1}, . . ., v_{m}$ as a basis for $V$ .

Notice if $v_{1} \in null (T)$ then that $T (v_{1}) = 0$ so we can start with any vector $w_{1} \in W$ then extend that to a basis $w_{1}, . . ., w_{n}$ for a basis of $W$ .

If $v_{1} \notin null (T)$ then $T (v_{1}) = w_{1}$ for some $w_{1} \in W$ . Again, extend this to a basis for $W$ .

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#9

Linear combination of columns

Suppose $A$ is an $m \times n$ matrix and $c = [\begin{matrix} c_{1} \\ ⋮ \\ c_{n} \end{matrix}]$ is an $n \times 1$ matrix. Then:

A c = c_{1} A_{\cdot, 1} + \dots + c_{n} A_{\cdot, n}

In other words, $A c$ is a linear combination of the columns of $A$ , with the scalars that multiply the columns coming from $c$ .

Proof
We have $A_{m \times n}$ and $C_{n \times 1} = c$ so then by the definition of matrix algebra we get:

\begin{aligned} (A C)_{j, 1} & = \sum_{r = 1}^{n} A_{j, r} C_{r, 1} \\ = \sum_{r = 1}^{n} A_{j, r} c_{r} \\ = {(\sum_{r = 1}^{n} A_{j, r} c_{r})}_{j, 1} \\ = c_{1} A_{\cdot, 1} + \dots + c_{n} A_{\cdot, n} \end{aligned}

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#10

Theorem

Suppose $A_{m \times n}$ and $C_{n \times p}$ . Then:

(A C)_{j, \cdot} = A_{j, \cdot} C

for $1 \leq j \leq m$ .

Proof

\begin{aligned} (A C)_{j, \cdot} & = \sum_{r = 1}^{n} A_{j, r} C_{r, 1}, . . ., A_{j, r} C_{r, p} \\ = A_{j, \cdot} C \end{aligned}

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# 12

Theorem

There's an example of $2 \times 2$ matrices that shows that matrix multiplication isn't commutative.

Proof

[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] \neq [\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}]

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