Theorem 7:
Let $A$ be an $n \times n$ matrix whose distinct eigenvalues are
$\{1,…, p\}$.a. For $1 \leq k \leq p$, the dimension of the eigenspace for $k$ is
less than or equal to the multiplicity of the eigenvalue $k$.b. The matrix $A$ is diagonalizable if and only if the sum of the
dimensions of the eigenspaces equals $n$, and this happens if and only
if (i) the characteristic polynomial factors completely into linear
factors and (ii) the dimension of the eigenspace for each $k$ equals
the multiplicity of $k$.c. If $A$ is diagonalizable and $B_k$ is a basis for the eigenspace
corresponding to $k$ for each $k$, then the total collection of
vectors in the sets $\{B_1,…,B_p\}$ forms an eigenvector basis for
$\mathbb{R^n}$.
Diagonalize the following matrix, if possible.
$A=\begin{bmatrix}5&0&0&0\\0&5&0&0\\1&4&-3&0\\-1&-2&0&-3\end{bmatrix}$.
Solution: Since $A$ is a triangular matrix, the eigenvalues are $5$ and $3$, each with multiplicity $2$.
Basis for $\lambda=5$: $v_1=\begin{bmatrix}-8\\4\\1\\0\end{bmatrix}$ and $v_2=\begin{bmatrix}-16\\4\\0\\1\end{bmatrix}$
Basis for $\lambda=-3$: $v_3=\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$ and $v_4=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$
The set $\{v_1,..v_4\}$ is linearly independent, by Theorem 7. So the matrix $P=[v_1,…v_4]$ is invertible.
Looking at Theorem 7, I can see how we can easily prove points a and b after we solved for the eigenvectors, but I don't see how Theorem 7 allowed us to say that the set $\{v_1,..v_4\}$ is linearly independent because it starts with "If A is diagonalizable", which is what we're trying to prove.
Best Answer
Theorem 7.b) states that $A$ is diagonalizable iff the sum of the dimensions of the eigenspaces is $n$. We have $n= 4$. Let $E_\lambda$ be the eigenspace for an eigenvalue $\lambda$. Since $\{v_1, v_2\}$ forms a basis for $\lambda = 5$ we have $\dim(E_5) = 2$ and similarly $\dim(E_{-3}) = 2$ and we have $\dim(E_5) + \dim(E_{-3}) = 4 = n$, thus by Theorem 7.b) $A$ is diagonalizable and therefore by Theorem 7.c) the set $\{v_1, v_2 ,v_3 ,v_4\}$ is linearly independent.