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.


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.