User:J D Bowen/Math725 HW7

1) Let $$(\mathbb{F}^n, S) \ $$ be $$\mathbb{F}^n \ $$ with the standard basis, and let $$(\mathbb{F}^n, B) \ $$ be the same vector space with basis $$B \ $$. We know that $$N(S)=B \iff \ $$ the columns of $$N \ $$ are the vectors of $$B \ $$.

Then $$T = \ $$ multiplication by $$M \ $$, $$N:(\mathbb{F}^n, S)\to (\mathbb{F}^n, B), \ T:(\mathbb{F}^n, B)\to (\mathbb{F}^n, B) \ $$ means

$$\mathfrak{M}_B^B (T) = N^{-1}TN \ $$.

2)

a) Let $$A \ $$ be an $$n\times n \ $$ diagonal matrix over $$\mathbb{F} \ $$. This represents a linear operator over $$\mathbb{F}^n \ $$. Since it is diagonal, there exists constants $$c_i \ $$ such that $$a_{ij}=c_i \delta_{ij} \ $$, and the eigenvalues are precisely the $$c_i \ $$, since they are the roots of $$\Pi (c_i-\lambda)=0 \ $$.

Let $$\vec{c}_i = (0,\dots,\underbrace{c_i}_{i^{th} \ \text{position}},0,\dots,0)^t \ $$. Then note that

$$A\vec{c}_i=(0,\dots,c_i^2,0,\dots,0)^t= c_i (0,\dots,c_i,0,\dots,0)^t=c_i\vec{c}_i \ $$

and so the $$\vec{c}_i \ $$ are eigenvectors. Since the are necessarily linearly independent, and since there are precisely $$n \ $$ of them, they span $$\mathbb{F}^n \ $$, the domain.

b)

We have the $$\Leftarrow \ $$ from (a). So $$V^n \ $$ is an n-dimensional vector space and suppose $$T:V^n\to V^n \ $$ is a diagonalizable operator. Then there is a set $$B=\left\{{\vec{v}_1, \dots, \vec{v}_m}\right\} \ $$ of eigenvectors of $$T \ $$ that span the domain, $$V^n \ $$. Since $$T \ $$ can have at most $$n \ $$ linearly independent eigenvectors, and since they span the domain, $$B \ $$ must be a basis for $$V^n \ $$. Then $$\mathfrak{M}_B^B (T) \ $$ is just $$m_{ij}=c_i \delta_{ij} \ $$, where $$T\vec{v}_i=c_i\vec{v}_i \ $$. This is a diagonal form for the operator.

c)

Let $$E_i =\text{span}(\vec{v}_i) \ $$ be the eigenspace associated with $$\vec{v}_i \ $$. Since $$\Sigma E_i = \Sigma \text{span}(\vec{v}_i) = \text{span}(v_1,\dots,v_m) \ $$, we have $$\Sigma E_i = V \iff V=\text{span}(v_1,\dots,v_m) \ $$, and by part (b), we have $$T \ \text{diagonalizable} \iff V=\text{span}(v_1,\dots,v_m) \ $$, so the theorem follows.

d) If $$b=0 \ $$ the matrix is diagonal and we may apply part (a). So assume $$b\neq 0 \ $$.

Suppose $$a\neq c \ $$. Then the eigenvalues are precisely $$a,c \ $$. Some eigenvectors, then, are

$$\vec{v}=(1,0) \ $$ and $$\vec{u}=(1,(c-a)/b) \ $$

Since $$c\neq a \ $$, the second term of $$\vec{u} \ $$ is not zero, and so the two vectors are linearly independent and thus span the space; hence, the matrix is diagonalizable.

Now suppose that the matrix is diagonalizable. Observe that as before the eigenvalues are $$a,c \ $$, and so for the first eigenvector $$\vec{v}=(v_1,v_2) \ $$, we have

$$av_1+bv_2=av_1, \ cv_2=av_2 \implies v_2 =0\implies (1,0) \ \text{is an eigenvector} \ $$.

For the second eigenvector, we have

$$au_1+bu_2=cu_1 \implies bu_2=(c-a)u_1 \ $$.

Since we know $$\vec{u} \ $$ is linearly independent from $$\vec{v} \ $$, we must have either $$(0,x) \ $$ or $$(1,x)\and x\neq 0 \ $$ as possible vectors for $$\vec{u} \ $$. If we suppose $$u_1=0 \ $$, then $$b=0 \ $$. If we suppose $$u_1=1 \ $$, then we have $$u_2=(c-a)/b\neq 0 \implies c\neq a \ $$. Therefore $$T \ $$ diagonalizable implies either $$a\neq c \ \or b=0 \ $$.

3) Let $$M, N, D \ $$ be $$n\times n \ $$ matrices, with $$D \ $$ diagonal. Then we can write $$d_{ij}=d_j \delta_{ij} \ $$.

Suppose $$MN=ND \ $$. For each column $$\vec{n}_j \ $$ of $$N \ $$and each $$d_j \ $$, this is equivalent to stating $$M\vec{n}_j = \vec{n}_j d_j \ $$, which is equivalent to stating that the $$\vec{n}_j \ $$ are eigenvectors of $$M \ $$ with eigenvalue $$d_j \ $$.

Next observe that if $$N^{-1} \ $$ is invertible, then kernel of $$N \ $$ is $$0 \ $$ and so the columns span the space. Since the columns are the eigenvectors of $$M \ $$, the eigenvectors of $$M \ $$ span the space, and so $$M \ $$ is diagonalizable.

4) Suppose $$\pi_1 : V\to W_1, \pi_2:V\to W_2 \ $$ are surjective linear maps with the same kernel.  Observe $$\text{dim}(\text{ker}(\pi_1)+\text{dim}(W_1)=\text{dim}(V)=\text{dim}(\text{ker}(\pi_2)+\text{dim}(W_2) \implies \text{dim}(W_1)=\text{dim}(W_2) \ $$.

Define $$T:W_1\to W_2 \ $$ as $$T(x)=\pi_2\left({\frac{\pi_1^{-1}(x)}{\text{ker}(\pi_1)} }\right) \ $$. Then $$(T\circ \pi_1)(x)= \pi_2(\pi_1^{-1}(\pi_1(x)))=\pi_2(x) \ $$.

Since vectors space are groups under addition, the first isomorphism theorem guarantees this is a bijection.