5.B: Eigenvalues/Vectors and UT Matrices

14

We'll give an example of an operator who's matrix with respect to some basis contains only 0's on the diagonal, while the operator is still invertible. By the question's own admittance, we should consider non-U.T matrices for this.

Consider:

Here all the diagonals are 0's. For the sake of simplicity let $V={\mathbb{R}}^2$, and $\beta$ is the standard basis of $\hat{i},\hat{j}$. Then since $M(T)$ is as described above then:

So define $T(v)$ for $v=(v_1,v_2)\in {\mathbb{R}}^2$ as:

notice that $M(T)$ is as described above, but the inverse matrix $M(T{)}^{-1}$ I claim is:

Note that I can show this because:

and I can show that:

since if I claim this I can verify that:

and:

so then $T$ is invertible.

15

For simplicity let $V={\mathbb{R}}^2$ and let $\beta$ be the standard basis. Consider the matrix:

Here the entries in the diagonal are all non-zero, but $T$ is not invertible as we'll show. Since $T\in \mathcal{L}(V)$ we need to show that $T$ isn't surjective or injective (just one). We'll do injectivity. To show it's not injective, we need to find two vectors $v,u\in V$ such that $Tu=Tv$ but $u\ne v$. Notice first that:

Hence, choose $u=(1,0)$ and $v=(0,-1)$. By the above showing, notice that $Tu=(-1,1)$ and:

but $u\ne v$, showing that $T$ isn't invertible.

16

We'll prove the above lemma using the below lemma:

Proof
Let $T\in \mathcal{L}(V)$ be arbitrary, where $V$ is:

• finite-dimensional: $\text{dim}(V)=n$;
• non-zero: $n\ne 0$;
• a complex vector space

Consider the transformation $\tau :{\mathcal{P}}_n(\mathbb{C})\to V$, where $\tau$ is defined by:

We should show that $\tau$ is itself linear. Let $p,q\in {\mathcal{P}}_n(\mathbb{C})$ be arbitrary. Notice that for additivity:

and similarly for homogeneity:

thus $\tau$ is linear. Notice that $\text{dim}(V)=n<n+1=\text{dim}({\mathcal{P}}_n(\mathbb{C}))$ so then we can use the aforementioned lemma to say that $\tau$ isn't injective. Thus, there's some vectors $p,q\in {\mathcal{P}}_n(\mathbb{C})$ such that:

Thus implying that a new polynomial $r=p-q$ is where:

where notice that $p\ne q$ so $r\ne 0$. Namely, notice that $r$ is a polynomial of degree $n$ or less. Denote $\text{deg}(r)=m\le n$. Then:

But then:

as we defined $\tau$ above. Therefore:

But then clearly $v$ is an eigenvector for at least one $(T-{\lambda }_{i}I)$ where $i\in \{1,...,m\}$. So then $v$ has some eigenvalue ${\lambda }_i$ associated with the transformation $T$, completing the theorem.
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20

Proof
For simplicity, let $\text{dim}(V)=n$. Fix $k=1,...,\text{dim}(V)$ as arbitrary. Let $v_1,...,v_n$ be the basis vectors that spans $V$.

Since $V$ is a finite-dimensional complex vector space and $T\in \mathcal{L}(V)$, then $T$ has an upper triangular matrix with respect to some basis of $V$, which we use as $\beta$ where $\beta =\{v_1,...,v_n\}$. As such, then the matrix of $M(T)$ with respect to $v_1,...,v_n$ is U.T.

But this is equivalent to saying that $\text{span}(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$. As such, for whatever $k$ we've fixed, have:

as per the reasoning above, then $U$ is invariant under $T$ because of our conditions under $V$, so then we can always find $U$ for each $k$ we've chosen. Furthermore, clearly $U$ is $k$-dimensional as $v_1,...,v_k$ are all LI, so then any $u\in U$ must be composed of these LI vectors, namely $k$ of them.
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5.C: Eigenspaces and Diagonal Matrices

1

Proof
Let $T\in \mathcal{L}(V)$ be diagonalizable. As such, then $V$ has a basis consisting of eigenvectors of $T$, we denote as $v_1,...,v_n$ (where $\text{dim}(V)=n$ is implicitly defined). As such, consider each $T{v}_j$ for any $j$. Since each $v_j$ is an eigenvector of $T$, we always get:

Now, consider the cases where ${\lambda }_j=0$ or not. If ${\lambda }_j=0$, then $T{v}_j=\overrightarrow{0}$ so $v_j\in \text{null}(T)$. Notice though that $v_j\notin \text{range}(T)$ since $v_j$ is an eigenvector, so the only vector where $Tw=v_j$ would be $\frac{1}{{\lambda }_j}{v}_j$ since:

But that would contradict ${\lambda }_j=0$.

If instead ${\lambda }_j\ne 0$ then $T{v}_j={\lambda }_j{v}_j$ so then clearly $v_j\in \text{range}(T)=\{Tv:v\in V\}$. Notice though that $T{v}_j\ne \overrightarrow{0}$ since both ${\lambda }_j\ne 0$ as well as $v_j\ne 0$ since it's an eigenvector.

Since $v_1,...,v_n$ is a basis and $T$ is linear, then any vector $v$ where $v$ is in the nullspace cannot be in the range, and vice versa, of $T$. As such, then both sets are disjoint. It's clear that $V=\text{null}(T)+\text{range}(T)$ by the FTOLM, but furthermore we know that since they're disjoint subsets whose sum makes up $V$ that then this must be a direct sum.
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3

Proof
It's clear that if (1) is true then (2, 3) are both true via the definition of a direct sum as well as Chapter 1 - Vector Spaces#^038942.

Suppose (2). Notice if we prove (3) then (1) comes for free. By the FTOLM then:

But since $\text{null}(T),\text{range}(T)$ are subspaces of $V$, then we also have it that:

Thus subtracting both equations, it's clear that $\text{dim}(\text{null}(T)\cap \text{range}(T))=0$ so then $\text{null}(T)\cap \text{range}(T)=\{\overrightarrow{0}\}$, proving (3), thus also getting (1) as well.

Suppose (3). Notice that if we prove (2) then (1) comes for free. Since $\text{null}(T)$ and $\text{range}(T)$ are subspaces of $V$ then:

But via our supposition then we know that the dimension of the negative term is 0, so:

As such, then we can represent any vector $v\in V$ by a set of basis vectors $u_1,...,u_m,w_1,...,w_k$ where $m$ is the dimension of the null space, and $k$ is the dimension of the range (of $T$). Thus:

where $u\in \text{null}(T)$ and $w\in \text{range}(T)$ so then since $v$ was arbitrary it follows that $V$ is at least a normal sum of the null-space and range (of $T$).
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6

Proof
Suppose the suppositions in the theorem. For clarity, there's ${\lambda }_1,...,{\lambda }_n$ distinct eigenvalues for $T$ and thus has corresponding eigenvectors $v_1,...,v_n$. Furthermore, since $S$ has the same eigenvectors as $T$, then $v_1,...,v_n$ are eigenvectors of $S$. We may have different $\lambda$'s so then these have corresponding eigenvalues ${\lambda }_1^{\prime },...,{\lambda }_n^{\prime }$.

Note here that since we have $v_1,...,v_n$ that these can form a basis of $V$, as there's $n$ vectors and they are all LI with each other (since they're eigenvectors, otherwise they'd share $\lambda$'s and thus $\lambda$ wouldn't be distinct). And note that for each $j$ that $T{v}_j={\lambda }_j{v}_j$ and furthermore $S{v}_j={\lambda }_j^{\prime }{v}_j$.

Consider any vector $v\in V$. It can be composed into a linear combination of it's basis vectors:

Thus notice that:

and likewise:

Thus since $v$ was arbitrary it follows that $ST=TS$.
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7

Proof
Since $M(T)=A$ is a diagonal matrix, then

where ${\lambda }_i$ may or may not be distinct. Clearly since $A$ is $n\times n$ then:

where we take ${\lambda }_1,...,{\lambda }_n$ and consider just the distinct eigenvalues ${\lambda }_1,...,{\lambda }_m$ eigenvalues.

Let's prove this via induction over $m\le n$., where we consider the arbitrary $\lambda \in \{{\lambda }_1,...,{\lambda }_m\}$. Consider $m=1$. Then:

This implies that $\lambda ={\lambda }_1$ is the only distinct eigenvalue from ${\lambda }_1,...,{\lambda }_n$, so then:

and since ${\lambda }_1$ appears $n$ times (since $A$ is $n\times n$), then that's the same as $\text{dim}({\lambda }_1,T)$ as we needed.

Now suppose that the theorem holds for any $k<m$. Consider the $k$-th case. Here we have:

Notice that since the $k-1$ case is true, then if $\lambda \in \{{\lambda }_1,...,{\lambda }_{k-1}\}$ then by the inductive hypothesis then we have $\lambda$ on the diagonal $\text{dim}(E(\lambda ,T))$ times as expected. If instead $\lambda ={\lambda }_k$ then we know that:

clearly we know that $\lambda \ne {\lambda }_i$ for $i=1,...,k-1$, so then by the inductive hypotheses for each ${\lambda }_i$ where ${\lambda }_i\ne {\lambda }_k$ that we will get $\text{dim}({\lambda }_i,T)$ times on the diagonal for $A$. As such, if a total of $n$ times will appear for all $\lambda$'s and we subtract the other $\text{dim}({\lambda }_i,T)$ times, then we get the equality above, showing that therefore for ${\lambda }_k$ we expect an appearance of $\text{dim}({\lambda }_k,T)$ times, as expected.

Thus, via the principle of mathematical induction, the theorem holds.
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8

Proof
Notice that $\text{dim}(V)=5$ in this case, and since we have the dimension above then:

Notice that $\text{dim}(E(\lambda ,T))\ne 0$ since if that's not the case then we have $E(\lambda ,T)=\{\overrightarrow{0}\}$ which is a contradiction as $\overrightarrow{0}$ itself isn't an eigenvector. As such, then we must have it that:

for some eigenvalue $\lambda \in \mathbb{F}$. As such, we have one other eigenvalue $\lambda \ne 8$ for $T$, with some eigenvector $v\in {\mathbb{F}}^5$.

Now assume for contradiction that $T-2I$ and $T-6I$ are both not invertible. That means that both $2,6$ are eigenvalues of $T$. But that's a clear contradiction since if that's the case then we have it that instead:

but clearly since none of these dimensions can be 0, then we cannot share the 1 across all values of the dimensions of our eigenspaces above. Hence, then we must have a contradiction, so then the opposite of our assumption is true. Hence, $T-2I$ or $T-6I$ must be invertible.
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10

Proof

First, we'll prove a nice lemma, that $E({\lambda }_i,T)\subseteq \text{range}(T)$, as this will be useful later on. Let $v\in E({\lambda }_i,T)$ be arbitrary. Since ${\lambda }_i\ne 0$ (see the theorem) then notice that we can say that $Tv={\lambda }_{i}v\Rightarrow T({\lambda }_i^{-1}v)=v$ so then by definition then $v\in \text{range}(T)$. So no matter what $v\in \text{range}(T)$. Thus, we've proved our lemma.

Now consider:

which is true via a quick inductive proof, using our lemma as our base case. Notice then that using the idea of the dimension of direct sums gives the left side, and it being a subset implies the $\le$ part:

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11

Proof
Notice that:

and:

thus then if ${\beta }^{\prime }$ is the basis respective to these new vectors then:

as expected.
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12

Proof
Notice that since $\text{dim}({\mathbb{F}}^3)=3$ then, if $e_1,e_2,e_3$ are the basis w.r.t. $T$ and $e_1^{\prime },e_2^{\prime },e_3^{\prime }$ is the basis w.r.t. $R$ then we know that:

and similarly:

Choose $S$ such that $S{e}_i^{\prime }=e_i$ for $i=1,2,3$. Namely:

Notice then that $S^{-1}$ exists since:

and furthermore, $S\in \mathcal{L}({\mathbb{F}}^3)$ by the similar properties for our bases.

Since $e_i^{\prime }$'s form a basis for our vector space, then for any $v\in {\mathbb{F}}^3$ then:

Thus:

Thus $S^{-1}TS=R$.
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14

Proof
The basis we are respective to doesn't matter, other than that there is some basis $e_1,e_2,e_3$ as we have $\text{dim}({\mathbb{C}}^3)=3$.

We'll want to find eigenvectors for our $T$ such that the sum of the dimensions of our eigenspaces per $\lambda$ doesn't add up to 3. We can just just force one eigenvector to occur per $\lambda$. Hence, we have:

Notice here we have $e_1$ is an eigenvector for $\lambda =6$ and similarly we have $\lambda =7$ with eigenvector $e_2$. Furthermore, notice that if $\beta =\{e_1,e_2,e_3\}$ then:

Notice that this is what we want! That's because we are always forced to have an UT matrix (and hence the diagonal tells what $\lambda$'s there are), so we have to use $6$ or $7$ in there on the diagonal, and we must have other vector components since if not then $e_3$ is an eigenvector of eigenvalue $\lambda =6$ or $\lambda =7$.

We can show that if $v\in V$ then $v={\alpha }_1{e}_1+{\alpha }_2{e}_2+{\alpha }_3{e}_3$ and then:

Creates equations:

Notice if $\lambda =6$ then (3) is satisfied (${\alpha }_3$ is arbitrary), and from (2) we have ${\alpha }_2=0$, and from (1) we have ${\alpha }_1=0$. Thus, any multiple of $e_1$ (${\alpha }_1$ is now arbitrary) is an eigenvector with that eigenvalue.

A similar process shows that if $\lambda =7$ then any multiple of $e_2$ is an eigenvector.

If $\lambda \ne 6,7$ then (2,3) force use to have it that ${\alpha }_2,{\alpha }_3=0$. Hence (1) becomes:

where since $\lambda \ne 6$ then we have it that ${\alpha }_1=0$ so we'd have the zero vector. Hence, there's no other eigenvalues or vectors.

As such, then $\text{dim}(E(6,T))=1=\text{dim}(E(7,T))$. Their sum is less than the dimension of ${\mathbb{F}}^3$, hence $T$ must not be diagonalizable.
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15

Proof
We claim that $(17,\sqrt{5},2\pi )\in {\mathbb{C}}^3$ is this vector. Notice that if $\lambda =8$ was an eigenvalue for $T$, then $T$ would have 3 distinct $\lambda$'s, and since $\text{dim}({\mathbb{C}}^3)=3$ then that would imply that $T$ is diagonalizable which is a contradiction. Thus, $\lambda =8$ must not be an eigenvalue.

As a result, then $T-8I$ must be surjective (if it wasn't, then since $T\in \mathcal{L}({\mathbb{C}}^3)$ then it would imply that $8$ is an eigenvalue which is a contradiction), so then there is some $(x,y,z)\in {\mathbb{C}}^3$ is non-zero such that $(T-8I)(x,y,z)=(17,\sqrt{5},2\pi )$. Then:

completing the proof.
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