Determine the span for an eigenspace

diagonalizationlinear algebramatrices

Currently, I am trying to solve the following exercise.
However, it is not clear to me how we can determine the spans by the given matrices.

I have tried to determine the null space for each eigenspace before determining the span, but I'm not getting the same answer as described in the solutions.

My answers:

$$
E_1 (\alpha =0)=
\begin{bmatrix}
1 \\
1 \\
0 \\
\end{bmatrix}\\
E_2(\alpha = 1)= \begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix}\\
E_1(\alpha=2)= \begin{bmatrix}
1 \\
0 \\
1 \\
\end{bmatrix}
$$

I've got the same eigenvalues and alpha values as in the solutions. I understand how I can compute this.

As you can see, I am not calculating the eigenspaces correctly, but I don't know where I made the mistakes.

Could someone explain to me how I can compute the eigenspaces from the matrices?

Thanks!

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Best Answer

Let's focus on the case where $\alpha = 0$. We want to compute the eigenspace associated with $\lambda = 1$. To do this, we calculate the nullspace of the matrix $$ M = A - \lambda I = \pmatrix{1 & \alpha & 0\\ 0 & \alpha + 1 & 1\\ 0 & 0 & 3 - \alpha} - \lambda \pmatrix{1&0&0\\0&1&0\\0&0&1} \\ = \pmatrix{1 & (0) & 0\\ 0 & (0) + 1 & 1\\ 0 & 0 & 3 - (0)} - (1)\pmatrix{1&0&0\\0&1&0\\0&0&1} \\ = \pmatrix{0&0&0\\0&0&1\\0&0&2}. $$ The nullspace of this matrix is the set of solutions to the equation $$ Mx = 0 \implies \pmatrix{0&0&0\\0&0&1\\0&0&2}\pmatrix{x_1\\x_2\\x_3} = 0 \\ \implies \begin{cases} 0=0\\ x_3 = 0\\ 2x_3 = 0. \end{cases} $$ There are a few ways to get a basis for the solution space, but I think that the most intuitive approach here is to note that $x$ will be a solution iff it is of the form $x = (x_1,x_2,0)$ (i.e. we only need to have $x_3=0$). We can express this as $$ x = x_1\pmatrix{1\\0\\0} + x_2 \pmatrix{0\\1\\0}. $$ Because the eigenspace $E_1(A)$ consists of all vectors of this form, we may state that $$ E_1(A) = \operatorname{span}\left\{\pmatrix{1\\0\\0},\pmatrix{0\\1\\0} \right\}. $$


Regarding the last eigenspace: we note that for $\lambda = 1, \alpha = 2,$ $$ A - \lambda I = \pmatrix{0&2&0\\0&2&1\\0&0&0}. $$ The corresponding system of equations is $$ 2x_2 = 0, \quad 2x_2 + x_3 = 0. $$ By plugging the first equation into the second, we come to the conclusion that these equations imply that $x_2 = x_3 = 0$. Thus, every vector can be written in the form $$ x = \pmatrix{x_1\\0\\0} = x_1 \pmatrix{1\\0\\0}, $$ which is to say that the eigenspace is the span of the vector $(1,0,0)$.

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