# Linear Algebra - seemingly incorrect result when looking for a basis

For the following matrix $$A = \begin{pmatrix} 3 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ -2 & -2 & 1 & 0 \\ -9 & -9 & 0 & -3 \end{pmatrix}$$

Find a basis for the eigenspace $E_{\lambda}(A)$ of each eigen value.

First step is to find the characteristic polynomial to find the eigenvalues. $$\det(A- I\lambda) = 0 \implies \lambda ^4 - \lambda^3 - 7 \lambda^2 + 13 \lambda -6 = 0$$ $$\lambda = 1 \lor \lambda = 2$$ Now that we have the eigen values, to find a basis for $E_{\lambda}(A)$ for the eigen values $\lambda$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda I)\mathbf{v} = 0$:

$$(A - 2I)\mathbf{v} = \begin{pmatrix} 1 & 1 & 0 & 0 \\ -2 & -2 & 0 & 0 \\ -2 & -2 & -1 & 0 \\ -9 & -9 & 0 & -5 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \implies \begin{cases} v_1 + v_2 = 0\\ -2v_1 -2v_2 = 0 \\ -2v_1 - 2v_2 -v_3 = 0 \\ -9v_1 -9v_2 - 5v_4 = 0 \end{cases}$$ $\implies v_2 = -v_1 \land v_3 = 0 \land v_4 = 0$. So $(v_1, v_2, v_3, v_4) = (v_1, -v_1, 0, 0) = v_1(1, -1, 0, 0)$ so $(1, -1, 0, 0)$ is the basis.

However, this is apparently incorrect. What have I done wrong here?

Also, for the eigenvalue $\lambda = 1$ I do get the correct answer, so only the part above gives me problems.

• First, there is a typo in the characteristic polynomial: the coefficient of $\lambda^4$ should be $1$, not $-1$. Second, you're missing a root of that equation. Finally, while you have correctly calculated the eigenvector corresponding to $\lambda=2$, there are two other eigenvalues left to deal with. – rogerl Dec 5 '15 at 17:26

The characteristic polynomial is $$\det(A-\lambda I) = \det\begin{pmatrix} 3-\lambda & 1 & 0 & 0 \\ -2 & 0-\lambda & 0 & 0 \\ -2 & -2 & 1-\lambda & 0 \\ -9 & -9 & 0 & -3-\lambda \end{pmatrix}$$ This is not a triangular matrix, but it is block triangular, so its determinant is $$\det\begin{pmatrix} 3-\lambda & 1 \\ -2 & 0-\lambda \end{pmatrix} \det\begin{pmatrix} 1-\lambda & 0 \\ 0 & -3-\lambda \end{pmatrix}= (2-3\lambda+\lambda^2)(1-\lambda)(-3-\lambda)= (1-\lambda)^2(2-\lambda)(-3-\lambda)$$ The algebraic multiplicity of $2$ and $-3$ is one; for $1$ it is two.

The computation of the basis for the eigenspace relative to $2$ can be done via elimination: \begin{align} A-2I=\begin{pmatrix} 1 & 1 & 0 & 0 \\ -2 & -2 & 0 & 0 \\ -2 & -2 & -1 & 0 \\ -9 & -9 & 0 & -5 \end{pmatrix} &\to \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -5 \end{pmatrix} && \begin{aligned} R_2&\gets R_2+2R_1\\R_3&\gets R_2+2R_1\\R_4&\gets R_2+9R_1 \end{aligned} \\&\to \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -5 \\ 0 & 0 & 0 & 0 \end{pmatrix} &&\begin{aligned} R_2&\leftrightarrow R_3\\R_3&\leftrightarrow R_4 \end{aligned} \\&\to \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} && \begin{aligned}R_2&\gets -R_2\\R_3&\gets-\frac{1}{5}R_3\end{aligned} \end{align} Thus a basis is given by the single vector $$\begin{pmatrix}-1\\1\\0\\0\end{pmatrix}$$ (but also your vector is correct, of course).

For the eigenspace relative to $1$: \begin{align} \begin{pmatrix} 2 & 1 & 0 & 0 \\ -2 & -1 & 0 & 0 \\ -2 & -2 & 0 & 0 \\ -9 & -9 & 0 & -4 \end{pmatrix} &\to \begin{pmatrix} 1 & 1/2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & -9 & 0 & -4 \end{pmatrix} &&\begin{aligned} R_1&\gets\frac{1}{2}R_1\\ R_2&\gets R_2+2R_1\\ R_3&\gets R_3+2R_1\\ R_4&\gets R_4+9R_1 \end{aligned} \\&\to \begin{pmatrix} 1 & 1/2 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & -9 & 0 & -4 \\ 0 & 0 & 0 & 0 \end{pmatrix} &&\begin{aligned} R_2&\leftrightarrow R_3\\ R_3&\leftrightarrow R_4 \end{aligned} \\&\to \begin{pmatrix} 1 & 1/2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} &&\begin{aligned} R_2&\gets-R_2\\R_3&\gets R_3+9R_2\\R_3&\gets-\frac{1}{4}R_3 \end{aligned} \\&\to \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} &&R_1\gets R_1-\frac{1}{2}R_2 \end{align} so the rank is $3$ and the basis consists of a single vector $$\begin{pmatrix}0\\0\\1\\0\end{pmatrix}$$

You're missing an eigenvalue! I think you've made a typo in your characteristic polynomial, writing $-\lambda^4$ instead of $\lambda^4$. The characteristic polynomial factorizes to $(x-2)(x+3)(x-1)^2$, hence the eigenvalues are $-3$, $2$, and $1$. The vector you have found is indeed a basis for the 2-eigenspace, and you can easily to compute that the basis vectors of the eigenspaces of $-3$ and $1$ are

$\left( \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 0 \\ 0 \\ 1 \\ 0 \end{array} \right)$ respectively.

If you calculate $\det(A-\lambda I)$ by the first row (and all the minors too) you get $$\det(A-\lambda I)=(3-\lambda)\lambda(1-\lambda)(-3-\lambda)+2(1-\lambda)(-3-\lambda)=(3+\lambda)(1-\lambda)(\lambda^2-3\lambda+2) =(3+\lambda)(1-\lambda)(\lambda-1)(\lambda-2).$$ So the eigenvalues are $1$ (with multiplicity $2$), $2$, and $-3$.

The eigenvectors (and so the eigenspace) for $2$ you found correctly. You still need to find the eigenspaces for $1$ and for $-3$.