(a) $\text{det}(B) = \text{det}(P^{-1}) \text{ det}(A) \text{ det}(P)$

(b), (c) By rank-nullity theorem, it suffices to show (c) only.
For invertible $P$, it is clear that $\text{null}(PA) = \text{null}(A)$. Thus, $\text{nullity}(PA) = \text{nullity}(BP) = \text{nullity}(A)$. Again, $\text{nullity}(BP) = \text{nullity}(P^{-1}BP)$, done.

(d) Since $\text{tr}(AB) = \text{tr}(BA)$ for square matrices $A$ and $B$, $\text{tr}(B) = \text{tr}(P^{-1} A P) = \text{tr}(APP^{-1}) = \text{tr}(A)$.

(e) We should show $\text{det}(\lambda I - C) = \text{det}(\lambda I - A)$. Since

$$\begin{aligned} \lambda I - C &= \lambda I - P^{-1} A P \\ &= \lambda P^{-1} P - P^{-1} A P \\ &= P^{-1} (\lambda P - AP) \\ &= P^{-1} (\lambda I - A) P \end{aligned}$$

$\lambda I - C$ and $\lambda I - A$ are similar! So, their null spaces are also equal.

Suppose $A$ has $n$ linearly independent eigenvectors, $\mathbf{p}_1, \cdots, \mathbf{p}_n$. Then, construct

$$\begin{aligned} P = [\mathbf{p}_1 \; | \; \cdots \; | \; \mathbf{p}_n] \end{aligned}$$

which is invertible as it has full column rank. Let $\lambda_1, \cdots, \lambda_n$ be corresponding eigenvalues and $D = \text{diag}(\lambda_i)$. It is obvious: $A \mathbf{p}_i = \lambda_i \mathbf{p}_i \to AP = PD$.

Conversely, suppose $P^{-1} A P = D = \text{diag}(\lambda_1, \cdots, \lambda_n)$. Then,

$$\begin{aligned} AP &= A[ \mathbf{p}_1 \; | \; \cdots \; | \; \mathbf{p}_n ] = [ A\mathbf{p}_1 \; | \; \cdots \; | \; A\mathbf{p}_n ] \\ PD &= [ \lambda_1 \mathbf{p}_1 \; | \; \cdots \; | \; \lambda_n \mathbf{p}_n ] \end{aligned}$$

Thus, $A\mathbf{p}_i = \lambda_i \mathbf{p}_i$ for all $i = 1, \cdots, n$, and it means $\mathbf{p}_i$’s are eigenvectors of $A$ corresponding to $\lambda_i$. Since $P$ is invertible, column vectors of $P$ are linearly independent. Done.

It suffices to show: if $\mathbf{v}_1, \cdots, \mathbf{v}_n$ are eigenvectors of $A$ that correspond to distinct eigenvalues $\lambda_1, \cdots, \lambda_n$, then the set $\{ \mathbf{v}_1, \cdots, \mathbf{v}_n \}$ is linearly independent.

Suppose not. Let $j$ be the maximal $j$ such that $\mathbf{v}_1, \cdots, \mathbf{v}_j$ are independent. Then, there exists $c_i \; (1 \leq i \leq j)$ such that

$$\begin{aligned} \mathbf{v}_{j+1} = c_1 \mathbf{v}_1 + \cdots + c_{j} \mathbf{v}_{j} \end{aligned}$$

Multiply BHS by $A$:

$$\begin{aligned} \lambda_{j+1} \mathbf{v}_{j+1} &= c_1 \lambda_1 \mathbf{v}_1 + \cdots + c_{j} \lambda_j \mathbf{v}_{j} \\ &= c_1 \lambda_{j+1} \mathbf{v}_1 + \cdots + c_{j} \lambda_{j+1} \mathbf{v}_{j} \end{aligned}$$

Thus

$$\begin{aligned} \sum_{i=1}^j (\lambda_i - \lambda_{j+1}) c_i \mathbf{v}_i = \mathbf{0} \end{aligned}$$

and it implies $c_i = 0$ for all $1 \leq i \leq j$, and $\mathbf{v}_{j+1} = \mathbf{0}$: contradiction since we defined eigenvectors be nonzero.

Suppose geometric multiplicity of $\lambda_0$ be $k$. Let $\{ \mathbf{v}_1, \cdots, \mathbf{v}_k \}$ be a basis for the eigenspace of $\lambda_0$.

Then, we can extend it to a basis $\{ \mathbf{v}_1, \cdots, \mathbf{v}_k, \cdots, \mathbf{v}_n \}$ of a vector space $V$. Let

$$\begin{aligned} P=[\mathbf{v}_1 \; | \; \cdots \; | \; \mathbf{v}_k \; | \; \mathbf{v}_{k+1} \; | \; \cdots \; | \; \mathbf{v}_n ] = [B_1 \; | \; B_2] \end{aligned}$$

Note that $P$ is invertible as it has full column rank. So

$$\begin{aligned} AP &= [ \lambda_0 \mathbf{p}_1 \; | \; \cdots \; | \; \lambda_0 \mathbf{v}_k \; | \; AB_2] \\ \end{aligned}$$

And, we will construct a similar matrix $C$ with $A$ which has $k$ or more algebraic multiplicity of $\lambda_0$. Let

$$\begin{aligned} C=\left[\begin{array}{cc} \lambda_0 I_k & X \\ 0 & Y \end{array}\right] \end{aligned}$$

Then,

$$\begin{aligned} AP &= [ \lambda_0 \mathbf{p}_1 \; | \; \cdots \; | \; \lambda_0 \mathbf{v}_k \; | \; PZ] \\ \end{aligned}$$

where $Z = \left[\begin{array}{l} X \\ Y \end{array}\right]$

Since $P$ is invertible, we can set a specific $Z$ such that $Z = P^{-1} A B_2$, so we can say $AP = PC \to C = P^{-1} AP$. Note that $A$ and $C$ have the same characteristic polynomial, and $\text{det}(\lambda I - C) = (\lambda - \lambda_0)^k \text{det} (\lambda I_{n - k} - Y)$. Thus, algebraic multiplicity of $\lambda_0$ is greater than or equal to $k$.

Since the sum of algebraic multiplicity is $n$, it suffices to show (a) only.

Let $\lambda_1, \cdots, \lambda_k$ be the $k$ distinct eigenvalues of $A$. Let $E_1, \cdots, E_k$ be the corresponding eigenspaces, let $B_1, \cdots, B_k$ be any bases for these eigenspaces. Let $d_i = \text{dim}(B_i)$ be geometric multiplicity of $\lambda_i$ where $B_i$ is a basis of $E_i$.

Denote $B_i = \{ \mathbf{v}_{i1}, \cdots, \mathbf{v}_{i d_i} \}$. Set $B = \{ \mathbf{v}_{11}, \cdots, \mathbf{v}_{1 d_1}, \cdots, \mathbf{v}_{k1}, \cdots, \mathbf{v}_{k d_k} \}$. Since $\sum_{i=1}^k d_i = n$, $B$ is a set of $n$ linearly independent eigenvectors by $\mathbf{Thm\ 4.}$, which implies $A$ is diagonalizable.

Let $\lambda_0$ be an eigenvalue of $A$ and $k$ be geometric multiplicity of $\lambda_0$. Let $B_0 = \{ \mathbf{u}_1, \cdots, \mathbf{u}_k \}$ be an otrthonormal basis for the eigenspace of $\lambda_0$.

Then, we can extend it to an orthonormal basis $B = \{ \mathbf{u}_1, \cdots, \mathbf{u}_n \}$ for $\mathbb{R}^n$ by Gram-Schmidt process. Let $P$ be a matrix whose columns are $\mathbf{u}_i$. Then

$$\begin{align*} AP = P \left[\begin{array}{cc} \lambda_0 I_k & X \\ 0 & Y \end{array}\right] = PC. \end{align*}$$

Hence $C = P^{-1} A P = P^\top A P \to C^\top = P^\top A^\top P = C$, which implies $C$ is also symmetric and $X = 0$. Also, characteristic polynomials of $A$ and $C$ are same: $(\lambda - \lambda_0)^k \text{ det}(\lambda I_{n-k} - Y) = (\lambda - \lambda_0)^k p_Y (\lambda)$ as they are similar.

Finally, we claim that $p_Y (\lambda_0) \neq 0$ which suggests that the algebraic multiplicity of $\lambda_0$ is exactly $k$.

Suppose not. Then, there exists $\mathbf{y} \in \mathbb{R}^{n-k}$ such that $Y \mathbf{y} = \lambda_0 \mathbf{y}$. Let $\mathbf{z} = \left[\begin{array}{cc} \mathbf{0} \\ \mathbf{y} \end{array}\right]$ be the vector in $\mathbb{R}^n$.

$$C x=\left[\begin{array}{c:c} \lambda_0 I_{n-k} \\ \hdashline 0 & Y \end{array}\right]\left[\begin{array}{l} 0 \\ \hdashline \mathbf{y} \end{array}\right]=\left[\begin{array}{c} 0 \\ \hdashline \lambda_0 \mathbf{y} \end{array}\right]=\lambda_0 \mathbf{z}$$

So $\mathbf{z}$ is an eigenvector of $C$, and $P \mathbf{z}$ is eigenvector of $A$. Also, $\mathbf{e}_1, \cdots, \mathbf{e}_k$ are also eigenvectors of $C$. Thus, $\{ \mathbf{e}_1, \cdots, \mathbf{e}_k, \mathbf{z} \}$ are linearly independent.

That means $\{ P\mathbf{e}_1, \cdots, P\mathbf{e}_k, P\mathbf{z} \}$ are linearly independent, and it contradicts: geometric multiplicity of $\lambda_0$ is larger than $k$.

Thus, for any eigenvalue of $A$, the algebraic multiplicity is exactly same with the geometric multiplicity, it is done.