Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick

Question

The matrix is: $ \begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix} $

The solution says that

$ B\cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \\ 6\end{pmatrix}$

$ B\cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$

$ B\cdot \begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$

Thus the eigenvalues are $ \lambda_{1}=6,\lambda_{2}=0 $

My question is, how can I easily find $\begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$ and $\begin{pmatrix} 6 \\ 6 \\ 6\end{pmatrix}$?

Is there any way to see it "quickly"?

Here is a way to see it "quickly". $(0,0,0)$ usually is not an eigenvector. — Dietrich Burde, Oct 13 '22 at 10:58
If every row adds up to the same quantity $\lambda$, then $\lambda$ is an eigenvalue and $(1,1,\ldots,1)^t$ is a corresponding eigenvector. — Arturo Magidin, Oct 13 '22 at 16:08
The eigenvector is not $ (0,0,0)$. The eigenvectors are the ones on the RHS i.e$ (1,1,-1) $ and $ (3,0,-1)$. — Lelouch, Oct 13 '22 at 19:01

Lee Mosher · Answer 1 · 2022-10-13T16:06:01.037

One way to instantly see that $0$ is an eigenvalue is that the three columns are clearly not linearly independent --- in fact, they're all constant multiples of each other! Next, any dependency vector of those columns gives you an eigenvector for the eigenvalue $0$. For example, clearly the expression $$\text{column 1 + column 2 - column 3} $$ evaluates to the zero column vector. Putting the three coefficients $1, 1, -1$ of that expression into a column vector you therefore get the following eigenvector for the eigenvalue $0$ namely $$ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$$ I'm sure, without looking back at your own post, you can easily "see" another dependency vector to get another eigenvector for the eigenvalue $0$ which is linearly independent of the one just given.

For the remaining eigenvector, you could notice that every column has constant coefficients, and therefore the expression $$\text{column 1 + column 2 + column 3} $$ evaluates to a column vector that also has constant coefficients. Since the three coefficients $1, 1, 1$ of that expression are constant, putting those three coefficients into column vector therefore gives you an eigenvector \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} And since the actual sum of the columns is
\begin{pmatrix} 6 \\ 6 \\ 6 \end{pmatrix} you immediately conclude that $6$ is the eigenvalue.

score 1 · Accepted Answer · answered Oct 13 '22 at 18:56

It's easy to see that $0$ is a multiple eigenvalue with multiplicity two as we can exhibit two linearly independent vector of the kernel (since all columns are obviously a multiple of the first one). Then the last eigenvalue $\lambda$ can be found using the Trace as the sum of the diagonal elements is also equal to the sum of the eigenvalues counted with their multiplicity : $0 + 0 + \lambda = 1+2+3$. So the eigenvalues are $0$ and $6$.

score 1 · Answer 3 · answered Oct 15 '22 at 06:01

The range of the given matrix is spanned by one vector: $$ \left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right] $$ Therefore, any eigenvector (which must be non-zero by definition) must be a scalar multiple of this vector, or it must be in the null space of the given matrix. The above is an eigenvector of the given matrix. The null space is spanned by $$ \left[\begin{array}{r}2 \\ -1 \\ 0\end{array}\right],\left[\begin{array}{r}3 \\ 0 \\ -1 \end{array}\right] $$ You can easily rewrite this null space in terms of the vectors given in your statement of the problem.

Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick

3 Answers3