Example 5.37 (Eigenspace) of Linear Algebra Done Right, by Sheldon Axler

Question

The following is from Linear Algebra Done Right, by Sheldon Axler.

$5.37$ $\ $ $\ $ Example $$\pmatrix{8 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5}$$ is a diagonal matrix.

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$5.37$ $\ $ $\ $ Example $\ $ $\ $ Suppose the matric of an operator $T\in\mathcal{L}(V)$ with respect to a basis $v_1,v_2,v_3$ of $V$ is the matrix in Example $5.35$ above. Then $$E(8,T)=\operatorname{span}(v_1),\quad E(5,T)=\operatorname{span}(v_2,v_3).$$

I understand that, since this matrix is upper-triangular, the eigenvalues of the transformation are the elements of the diagonal: $8$ and $5$.

So the eigenspace of the transformation corresponding to the eigenvalue $\lambda$ is equivalent to the following null space: $null(T - \lambda I)$.

But this null space would just consist of all the eigenvectors, since $T\mathbf{v} = \lambda \mathbf{v} \rightarrow (T - \lambda I)\mathbf{v} = \mathbf{0}$, right?

But I still don't understand how the author came to the solutions $E(8, T) = \operatorname{span}(v_1)$ and $E(5, T) = \operatorname{span}(v_2, v_3)$?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

Note that$$T-8\operatorname{Id}=\begin{pmatrix}0&0&0\\0&-3&0\\0&0&-3\end{pmatrix}.$$Therefore$$T(\alpha_1v_1+\alpha_2v_2+\alpha_3v_3)=0\iff-3\alpha_2v_2-3\alpha_3v_3=0\iff\alpha_2=\alpha_3=0$$and so $\operatorname{null}(T-8\operatorname{Id})=\langle v_1\rangle$. The same argument applies to the other eigenvalue.

Answer 2

I understand that, since this matrix is upper-triangular

This holds for lower triangular matrices as well. You may remove the "upper".

But I still don't understand how the author came to the solutions $E(8, T) = span(v_1)$

If you subtract $8\operatorname{Id}$ from $T$, you obtain a matrix whose kernel is quite obviously generated by $v_1$.

Edit: the matrix you obtain sends $v_1$ to $0$ and acts as a non-zero multiple of the identity on the subspace $\left<v_2,v_3\right>$. Therefore, an element of the kernel must have no part in $\left<v_2,v_3\right>$ and lie entirely in $\left<v_1\right>$.