Chapter 4 - Polynomials

2

This is a false statement. Notice here that if we have $p(z)=z^m+1$ and $q(z)=-z^m$ then both $p,q\in U$ but $(p+q)(z)=z^m+1-z^m=1\notin U$ since it's degree is $-\mathrm{\infty }$. Hence, $U$ isn't closed under vector addition.

3

This is also a false statement. Let $p(z)=z^4+z^3$ and $q(z)=-z^4$. Then $(p+q)(z)=z^3\notin U$, showing that $U$ isn't closed under vector addition.

4

Proof
Consider if $m=n$. Then clearly there's a polynomial $p$ where:

coming from desiring $0=p({\lambda }_1)=\cdots =p({\lambda }_m)$. This step comes from the factorization of a polynomial over $\mathbb{C}$. Here $\text{deg}(p)=n$ as desired, and clearly $p$ doesn't have any other zeros.

Now consider if $m\ne n$. Then $m<n$. We need to add zeroes (to increase the degree) without adding "new" zeros. Hence use:

Hence we have found a $\text{deg}(p)=n$ polynomial where $p({\lambda }_1)=\cdots =p({\lambda }_m)=0$. As required.

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6

Proof
Consider $\to$, so suppose $p$ has $m$ distinct zeros ${\lambda }_1,...,{\lambda }_m$. Then:

Using the division algorithm, we can show that for any zero ${\lambda }_i$ above that:

and using the derivative on both sides:

Thus, notice that ${\lambda }_i$ cannot be shared between $p$ and $p^{\prime }$, since the above expression shows any ${\lambda }_i$ zero from $p$ must give a remainder polynomial $q(z)$ for $p^{\prime }$ unless $q(z)=0$, which would suggest $p=0$ and thus $p$ has 0 distinct zeroes and thus is its own contradiction. Therefore, $p$ and $p^{\prime }$ have different zeroes.

Now $\leftarrow$. Suppose that $p$ and $p^{\prime }$ have no zeroes in common. Assume for contradiction that $p$ can't have $m$ distinct zeroes. Since $p$ is a degree $m$ polynomial, then we must have at least $m$ zeroes, so therefore at least one zero ${\lambda }_i$ must be repeated. Namely:

where since ${\lambda }_i$ is repeated, it's must occur at least twice. Deriving it:

So then clearly both $p,p^{\prime }$ share the same zero ${\lambda }_i$, which contradicts our supposition from the beginning. Hence, our assumption was wrong, so $p$ must have $m$ distinct zeroes.
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7

Proof
Let $p\in \mathcal{P}(\mathbb{F})$ be and arbitrary odd degree real-coefficient polynomial. Assume for contradiction that $p$ only contains complex zeros, ie:

where it is assumed that $m$ is odd. Since any polynomial with real coefficients has pairs of zeroes $({\lambda }_i,\overline{{\lambda }_i})$ then we can, without loss of generality, say that ${\lambda }_2=\overline{{\lambda }_1},{\lambda }_4=\overline{{\lambda }_3},...$ so that:

But look! Since we have an odd number of ${\lambda }_i$'s, pairing them up this way gives one last ${\lambda }_{m/2}$ that has no conjugate pair, which contradicts all polynomial zeros having their conjugate pair being a zero. Hence, $p$ must contain a non-complex zero, namely a real one.
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Chapter 5.A: Invariants (Finish)

1

Suppose $T\in \mathcal{L}(V)$ and $U$ is a subspace of $V$.

a

Proof
Suppose $U\subset \text{null}(T)$. Let $u\in U$ be arbitrary. We'll need to show that $Tu\in U$ as well.

Notice that since $u\in U$, then since $U$ is a subset of the nullspace of $T$ then $u\in \text{null}(T)$, so then $Tu=\overrightarrow{0}\in U$. Hence, $U$ is an invariant under $T$.
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b

Proof
Let $u\in U$ be arbitrary. We'll need to show that $Tu\in U$. Note that since $Tu\in \text{range}(T)$ then $Tu\in U$ as required.
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2

Proof
Let $u\in \text{null}(S)$ be arbitrary. We'll need to show that $Tu\in \text{null}(S)$, meaning that $STu=\overrightarrow{0}$. Since $u\in \text{null}(S)$ then $Su=\overrightarrow{0}$. That means that:

since $ST=TS$. Hence, $Tu\in \text{null}(S)$ so then $\text{null}(S)$ is invariant under $T$.
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3

Proof
Let $u\in \text{range}(S)$ be arbitrary. We'll need to show that $Tu\in \text{range}(S)$. Since $u\in \text{range}(S)$ then there is some vector $u^{\prime }\in U$ such that $S{u}^{\prime }=u$. Then notice that:

so then since $T{u}^{\prime }$ is just some vector $v\in U$ then $Sv=Tu$ so then clearly $Tu\in \text{range}(S)$. Therefore, then $\text{range}(S)$ is invariant under $T$.
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4

Proof
Let $u\in U_1+\cdots +U_m$ be arbitrary. We'll need to show that $Tu\in U_1+\cdots +U_m$.

Since $u\in U_1+\cdots +U_m$ then:

where all $u_i\in U_i$ for $1\le i\le m$. Since $T$ is a linear transformation then:

notice that $T{u}_1\in U_1$ since $U_1$ is $u=u_1+\cdots +u_m$ is an invariant subspace of $V$ under $T$. In general, all $T{u}_i\in U_i$ for similar reasoning so then clearly:

so then $U_1+\cdots +U_m$ is invariant under $T$.
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5

Proof
Let $U$ be in the intersection of every collection of subspaces of $V$ invariant under $T$. Namely, $U=\bigcap _{i\in I}{U}_i$ where each $U_i$ is invariant under $T$. Let $u\in U$ be arbitrary. We want to show that $Tu\in U$ consequently.

Since $u\in U$ then for all $i\in I$ we have that $u\in U_i$. Since all $U_i$ are invariant under $T$, then $Tu\in U_i$ for all $i\in I$. Therefore, we have it that $Tu\in U$, showing $U$ is invariant under $T$.
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8

Define $T\in \mathcal{L}({\mathbb{F}}^2)$ by:

The eigenvalues and eigenvectors are found as follows. Let $\lambda$ be an eigenvalue for $T$. Then:

thus then:

so since $z\ne 0$ as $(z,w)\ne 0$ (notice if $z=0$ then that makes $w=0$) then ${\lambda }^2-1=0\Rightarrow \lambda =\pm 1$ so then the eigenvalues are $\boxed{{\displaystyle \lambda =\pm 1}}$.

Plugging back in to get our eigenvectors:

Solve for $w,z$ for both cases:

so have ${\overrightarrow{v}}_1=(1,1)$ associated with $\lambda =1$, and then:

so have ${\overrightarrow{v}}_2=(1,-1)$ associated with $\lambda =1$.

9

Define $T\in \mathcal{L}({\mathbb{F}}^3)$ by:

then we try to find the $\lambda$'s for this transformation:

equate coordinates:

So we have 3 equations and we have 3 unknowns. An easy option is if $\lambda =0$. Then:

Hence the vector $(1,0,0)$ is associated with eigenvalue $\lambda =0$.

If instead we consider the other obvious one $\lambda =5$ then:

Hence the vector $(0,0,1)$ is associated with the eigenvalue $\lambda =5$.

Notice that these are the only eigenvalues here. We could turn the system of equations into a matrix:

and solve using the characteristic polynomial to show this. But alternatively, we can see that the first equation will never be "reduced" to a true statement for any $\lambda$, so then we only had the two obvious choices above.

11

Define $T:\mathcal{P}(\mathbb{R})\to \mathcal{P}(\mathbb{R})$ by $Tp=p^{\prime }$. We'll find all eigenvalues and vectors for $T$.

We need to find when:

simplifying:

equating $x^n$ coefficients yields a system of equations:

Notice that if $\lambda =0$ then we get that all $a_i=0$ for $i\ge 1$ meaning $p=c+0x+\cdots =c$ which is a constant function. Hence all constant polynomials have $\lambda =0$ as an eigenvalue.

If instead we have $\lambda \ne 0$ then we have the simplified case looking at the last equation that $a_n=0$. Hence, then $n{a}_n=0$ so then $a_{n-1}=0$. Repeating, we get that $a_n,a_{n-1},\dots ,a_1,a_0=0$ so then $p$ is the zero function which isn't an eigenvector. Hence there's no eigenvectors for any $\lambda \ne 0$.

Overall, the only eigenvalue is $\lambda =0$ of eigenvector $p=c$ for any $c\in \mathbb{R}$.

12

Define $T\in \mathcal{L}({\mathcal{P}}_4(\mathbb{R}))$ by:

for all $x\in \mathbb{R}$. We'll find all eigenvalues and vectors of $T$.

simplifies to:

We can equate coefficients to get:

But look! Clearly if $\lambda =0$ then that forces all $a_i=0$ for $i\ge 1$, while $a_0\in \mathbb{R}$ so then $p=c$ which is any constant function. Hence, $\lambda =0$ corresponds to the constant function eigenvector.

If instead $\lambda \ne 0$ then this is how we can construct our eigenvalues. Notice if all $a_j=0$ for $j\ne k$ while $a_k\ne 0$ then we have true statements and have the remaining equation $k{a}_k=\lambda a_k$. Since $a_k\ne 0$ then we have (by dividing) that $k=\lambda$. Looking at the resulting polynomial, that gives:

Hence, in general any $p_k(x)$ (as defined above) is an eigenvector with $\lambda =k$ as it's eigenvalue.

14

Suppose $V=U\oplus W$, where $U,W$ are nonzero subspaces of $V$. Define $P\in \mathcal{L}(V)$ by $P(u+w)=u$ for $u\in U$ and $w\in W$. Find all eigenvalues and vectors of $P$.

Let $v\in V$ be arbitrary. Let $u_i\in U$ and $w_i\in W$. Consider:

But since we have a direct sum, we can't have $u_i=c{w}_i$ for any constant, since if that was the case then $v=u_i+w_i=u_i+c{u}_i=(1+c)u_i$ which isn't unique. Hence, we must have:

Which only happens if $\lambda =0,1$. If $\lambda =0$ then we get:

and thus $v=w_i$ which is a (possibly) non-zero vector. Hence, any $w\in W$ is an eigenvector with eigenvalue $\lambda =0$.

Similarly, if $\lambda =1$. Then we get:

so then $v=u_i$. Hence, any $u\in U$ is an eigenvector with eigenvalue $\lambda =1$.

15

Suppose $T\in \mathcal{L}(V)$ and $S\in \mathcal{L}(V)$. Suppose $S$ specifically is invertible.

a

Proof
Let $\lambda$ be an arbitrary eigenvalue from $T$, with corresponding $v\in V$ eigenvector. Thus:

Because $S$ is invertible, then $S$ is both injective and surjective. Surjectivity will come into play here, since that means there's some $v_o\in V$ where $S{v}_o=v$. Thus:

Thus $v_o$ is an eigenvector for any eigenvalue $\lambda$, for the transformation $S^{-1}TS$. Hence, since $\lambda$ was arbitrary, it holds for all eigenvalues that $\lambda$ is shared between $T$ and $S^{-1}TS$.

WLOG, the reverse direction holds a similar proof and shows that $S^{-1}TS$ eigenvalues are shared in $T$ similarly.
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b

Notice that prior that the eigenvector $v$ from $T$ are not the same for $S^{-1}TS$. Namely, the new eigenvector $v_o$ associated with $S^{-1}TS$ is $S^{-1}v=v_o$ for each eigenvector $v$ from $T$.

18

We'll show that the operator $T\in \mathcal{L}({\mathbb{C}}^{\mathrm{\infty }})$ defined by:

has no eigenvalues. Assume for contradiction that it does, at least one $\lambda$, with some unknown eigenvector $(z_1,...)$. Then:

We can equate coefficients to get:

But look at this. Clearly an option is $\lambda =0$. If that's the case then $z_1=z_2=\cdots =0$ so then $\overrightarrow{z}=\overrightarrow{0}$ which isn't a valid eigenvector. So instead consider that $\lambda \ne 0$. Then from the first equation we have it that $z_1=0$, but then from the second equation then $z_2=0$, and so on. This means that:

Hence then $\overrightarrow{z}=(0,0,...)=\overrightarrow{0}$ which again isn't a valid eigenvector. Hence, no $\lambda$ cases allow for eigenvectors to arise, so there are no valid $\lambda$.

19

Suppose $n$ is a positive integer and $T\in \mathcal{L}({\mathbb{F}}^n)$ is defined by:

so then $T$ is the operator whose matrix (w.r.t the standard basis) consists of all 1's. We'll find all eigenvalues and eigenvectors of $T$.

Consider when:

Equating rows of our vector, we have:

Namely, $x_1+\cdots +x_n=\lambda x_i$ for all $1\le i\le n$. That means that:

so then we have two possibilities. If $\lambda =0$, then we just require that $x_1+\cdots +x_n=0$, so then all vectors $v\in \{v\in {\mathbb{F}}^n:v_1+\cdots +v_n=0\}$ is a valid eigenvector with eigenvalue $\lambda =0$.

If instead we have $\lambda \ne 0$. Then the RHS of our chain of equations suggests that:

by dividing $\lambda$ from all sides. If $k=0$ we get a vector from before, so assume that $k\ne 0$. If that's the case, then:

Thus:

So the only $\lambda \ne 0$ occurs if $n=1$, and in that case of $T(x_1)=x_1$ then we just get the identity map. Hence, any $x\in {\mathbb{F}}^1$ is an eigenvector of $T$ with eigenvalue $\lambda =1$.

20

Consider $T\in \mathcal{L}({\mathbb{F}}^{\mathrm{\infty }})$ defined by:

We'll find all eigenvalues and eigenvectors of $T$:

Equating rows gives:

If $\lambda =0$ then we get that $z_2=z_3=\cdots =0$ so then we have $(z_1,0,0,...)$ is an eigenvector with eigenvalue $\lambda =0$.

If instead $\lambda \ne 0$ then notice that:

Notice in general that we can show that $z_n={\lambda }^{n-1}{z}_1$. We showed our base case, so let our inductive hypothesis reign for $n<k$, and we'll show the $k$-th case:

showing the $k$-th case. Notice if $z_1=0$ then all $z_i=0$ giving the zero vector which is invalid. So suppose $z_1\ne 0$. Then for any $\lambda \in \mathbb{F}\setminus \{0\}$ we can create an eigenvector ${\overrightarrow{z}}_{\lambda }$ defined as:

then this eigenvector has an associated eigenvalue of $\lambda \ne 0$. We can verify this:

Which works. As an example, if $z_1=1$ and $\lambda =\sqrt{2}$ then our eigenvector ${\overrightarrow{z}}_{\sqrt{2}}$ is:

as expected.

21

Suppose $T\in \mathcal{L}(V)$ is invertible.

a

Proof
Consider $\to$, so suppose $\lambda$ is an eigenvalue of $T$. Then there's some associated eigenvector $v\in V$. Then notice that:

thus $\frac{1}{\lambda }$ is an eigenvalue for $T^{-1}$.

WLOG, going in the reverse direction shows the $\leftarrow$ part.
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b

Proof
Let $\lambda$ be an eigenvalue for $T$ with eigenvector $v$. As shown above, we have the same eigenvalue $\frac{1}{\lambda }$ for $T^{-1}$. We need to show that the eigenvector associated with that value is still $v\in V$.

Notice that:

So then $\frac{1}{\lambda }$'s eigenvector for $T^{-1}$ is still $v$ as we needed. Since $\lambda$ was an arbitrary eigenvalue, then for all eigenvalues for $T$ we have sharing eigenvectors. This works in both directions.
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23

Proof
Let $\text{dim}(V)=n$. Let $\lambda$ be an eigenvalue of $ST$ with corresponding eigenvector $v\in V$. Notice that:

Thus then $\lambda$ is an eigenvalue for $TS$ with the corresponding eigenvector $Tv$. Since $\lambda$ was arbitrary, this holds true for all eigenvalues from $ST$ to $TS$. A similar proof by swapping $S$ with $T$ shows the reverse direction.
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29

Proof
Assume for contradiction that $T$ has $k+2$ or more eigenvalues. Hence, at least ${\lambda }_1,...,{\lambda }_{k+2}$ are all distinct eigenvalues of $T$. But then since $\text{dim}(\text{range}(T))=k$ then there's a basis $r_1,...,r_k$ for $\text{range}(T)$, so then $T{v}_1,...,T{v}_k$ is that same basis.

Notice in the best case that all $v_1,...,v_k$ are eigenvectors, so then ${\lambda }_1,...,{\lambda }_k$ are their associated eigenvalues. But we still have ${\lambda }_{k+1},{\lambda }_{k+2}$, so since this is the best case where we have the maximum number of eigenvectors for $\text{range}(T)$ then we require that ${\lambda }_{k+1}={\lambda }_{k+2}=0$ as the other vectors need to be send to the zero vector.

But look! ${\lambda }_{k+1}={\lambda }_{k+2}$, so they aren't distinct! This contradicts our supposition that they be distinct, so then we must have the opposite of the assumption. Thus $T$ has at most $k+1$ distinct eigenvalues.
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Chapter 5.B: Eigenvectors and UT Matrices

1

Suppose $T\in \mathcal{L}(V)$ and there is some positive integer $n$ such that $T^n=0$.

a

Proof
We can just show that:

Thus we've happened to find an invertible transformation $(I-T{)}^{-1}$ as defined above, and showed that it worked. Thus, it's the inverse matrix, showing $I-T$ is invertible.
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b

Usually we pull these ideas from properties of the real numbers, or other discrete math. Namely, we know that:

where since we don't have an infinite vector space we need:

where all $x^n,...,=0$ hence the condition of $T^n=0$ as if that's true then $T^{n+i}=(T^n{)}^i=0^i=0$ for all $i\ge 0$.

2

Proof
Let $\lambda$ be an eigenvalue of $T$. Then we know that:

where $v$ is $\lambda$'s associated eigenvector. Now, assume for contradiction that $\lambda \ne 2,3,4$ (all). Notice that:

Notice that if $(T-4I)v=\overrightarrow{0}$ that would be a contradiction as then $\lambda =4$ would be a valid eigenvalue. Hence, say $(T-4I)v=w\in V$ where $w$ is non-zero. Then:

In a similar way, if $(T-3I)w=\overrightarrow{0}$ then that would imply that $\lambda =3$ is an eigenvalue which is a contradiction. Hence, say $(T-2I)u=\overrightarrow{0}$ for nonzero $u\in V$. Then:

so then $\lambda =2$ is an eigenvalue is a contradiction. Hence, $\lambda =2,3,\text{or }4$.
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4

Proof
We can show this if $\text{null}(P)$ and $\text{range}(P)$ only contains the zero vector. Let $v\in V$ be any non-zero vector. Assume for contradiction that $v\in \text{null}(P)$ and $v\in \text{range}(P)$. Then $Pv=\overrightarrow{0}$ and further there is some vector $w\in V$ where $Pw=v$. Notice though that:

while:

so then:

Which contradicts $v$ being non-zero. Hence $\text{null}(P)$ is disjoint with $\text{range}(P)$ so then we get a direct sum. This is also a valid sum since, by the FTOLM:

as $\text{null}(P)$ and $\text{range}(P)$ are disjoint.
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5

Proof

First, we'll prove a small lemma, namely that $(ST{S}^{-1}{)}^n=S{T}^n{S}^{-1}$.

First, notice that $n=1$ case is satisfied. Suppose the lemma works for all cases under $k$. Consider the $k$-th case:

hence completing the proof for the lemma.

We know that, supposing $p(z)=a_0+a_{1}z+\cdots +a_n{z}^n$:

Completing the proof.
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7

Proof
Consider $\to$, so suppose $\lambda =9$ is an eigenvalue of $T^2$, so then:

for some $v\in V$. Notice that:

where $p(z)=z-3$ and $q(z)=z+3$. Notice that $p(z)q(z)=z^2-9$ so then:

thus:

now assume for contradiction that $3$ and $-3$ are not eigenvalues of $T$. Then notice that we must have $(T+3I)v\ne \overrightarrow{0}$ since if it did then $-3$ would be a valid eigenvalue of $T$. Hence, call this vector $w\in V$. Then:

So then $\lambda =3$ is an eigenvalue, which is also a contradiction. Hence, no matter what we get a contradiction, so then either $3$ or $-3$ are eigenvalues of $T$.

Now consider $\leftarrow$. Suppose $3$ or $-3$ are an eigenvalue of $T$. WLOG, suppose $\lambda =3$ is an eigenvalue. Then:

for some associated eigenvector $v\in V$. Then:

So then notice that $T^{2}v=9v$ so then $\lambda =9$ is an eigenvalue for $T^2$.

In a similar way if $-3$ was the eigenvalue, then:

so then $\lambda =9$ is still an eigenvalue for $T^2$.
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8

We want to find an example of $T\in \mathcal{L}({\mathbb{R}}^2)$ where $T^4=-1$. This suggest that:

Taking some inspiration from complex number's it's like we want to find the solutions to:

But notice that these won't work as standard eigenvalues, as we need to have real eigenvalue components for our vectors. But instead, this motivates us to use rotation as a way to model this transformation.

Let $T$ be the transformation that rotates a vector $\pi /4$ radians. Note that we can find the matrix under the standard basis as: