Without calculation, find one eigenvalue and two linearly independent eigenvectors of $A = \begin{bmatrix} 2 && 2 && 2 \\ 2 && 2 && 2 \\ 2 && 2 && 2 \end{bmatrix} $.

This matrix is non-invertible because its columns are linearly dependent. So the number $0$ is an eigenvalue of $A$. Eigenvectors for the eigenvalue $0$ are solutions of $Ax=0$ and therefore have entries that produce a linear independence relation among the columns of A. Any nonzero vector in ($\Bbb R^3$) whose entries sum to $0$ will work. Find any two vectors that are not multiples; for instance,

$\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix} $ and $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} $.

I do not understand how the first vector is obtained.

So far I've done.

$ \sim \begin{bmatrix} 1 && 1 && 1 \\ 0 && 0 && 0 \\ 0 && 0 && 0 \end{bmatrix} $

$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -x_2-x_3 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} $

How do you obtain $\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix} $?

Why is it that any nonzero vector in ($\Bbb R^3$) whose entries sum to $0$ will work?

  • $\begingroup$ "why is it that any nonzero vector whose entries sum to $0$ will work?" Try multiplying such a vector to it and see what happens. Since all entries in $A$ are equal, you could simply factor out a common $2$. "how do you obtain..." By setting $x_2=1$ and $x_3=-2$, your two vectors you found add up to the vector in question. Since it was stated to do this without calculation, several answers are possible. Your two vectors for example are also correct answers. $\endgroup$ – JMoravitz Oct 19 '15 at 1:36
  • $\begingroup$ Why must the entries sum to $0$? $\endgroup$ – user1766555 Oct 19 '15 at 1:40
  • $\begingroup$ "Without calculation"? That's impossible. Maybe they mean "without explicitly solving the characteristic polynomial and following matrix equation systems". $\endgroup$ – mathreadler Oct 19 '15 at 1:58
  • $\begingroup$ @mathreadler, no calculation is necessary to find all eigenvalues and eigenvectors of any multiple of the $Ones$ matrix beyond simple addition. $\endgroup$ – JMoravitz Oct 19 '15 at 2:00

"Why must the entries sum to $0$"

Let $A=\begin{bmatrix}a_{1,1}&a_{1,2}&a_{1,3}\\ a_{2,1}&a_{2,2}&a_{2,3}\\a_{3,1}&a_{3,2}&a_{3,3}\end{bmatrix}$ and $x = \begin{bmatrix} x_{1}\\x_2\\x_3\end{bmatrix}$

By definition of matrix product, you have

$Ax = \begin{bmatrix}a_{1,1}x_1 + a_{1,2}x_2 + a_{1,3}x_3\\a_{2,1}x_1 + a_{2,2}x_2 + a_{2,3}x_3\\a_{3,1}x_1 + a_{3,2}x_2 + a_{3,3}x_3\end{bmatrix}$

Note what happens when all entries of $A$ are identical and nonzero:

$\dots = \begin{bmatrix} a(x_1+x_2+x_3)\\a(x_1+x_2+x_3)\\a(x_1+x_2+x_3)\end{bmatrix} = a\begin{bmatrix}x_1+x_2+x_3\\x_1+x_2+x_3\\x_1+x_2+x_3\end{bmatrix}$

If this were to equal the zero vector (which is implied since these are eigenvectors for the eigenvalue of zero) then that means that every entry above must be zero. That is exactly the same as saying that $x_1+x_2+x_3=0$

On the other hand, supposing that $x_1+x_2+x_3=k\neq 0$, you would have $Ax = a\begin{bmatrix}k\\k\\k\end{bmatrix}\neq \overrightarrow{0}=0x$ and so $x$ would not be an eigenvector for the eigenvalue of zero.

In general, any nonzero multiple of the $Ones_{n\times n}$ matrix, say $a\cdot Ones_{n\times n}$ (matrix where all entries are $a$) will have an eigenvalue of zero and every eigenvector for zero will satisfy the relation that the sum of the entries is zero. We know that there are no others by a rank-nullity argument and that the remaining eigenvalue is equal to the trace (sum of the diagonal).

Furthermore, the other eigenvalue will necessarily be $a\cdot n$ with eigenvector $\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}$.

  • $\begingroup$ Why does it matter that the entries must sum to $0$ when $\lambda x = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $ due to lambda being $0$? $\endgroup$ – user1766555 Oct 19 '15 at 1:47
  • 1
    $\begingroup$ @user1766555 Suppose that they didn't sum to zero, but instead summed to $k$. Then going through the matrix multiplication, you see that $Ax = \begin{bmatrix}2k\\2k\\2k\end{bmatrix}\neq \overrightarrow{0}=0x$. This could also be seen directly from a rank-nullity argument and theorems about eigenvalues (namely that the sum of the eigenvalues must be equal to the sum of the diagonal). $\endgroup$ – JMoravitz Oct 19 '15 at 1:51
  • $\begingroup$ So due to $Ax = \begin{bmatrix}2k\\2k\\2k\end{bmatrix}\neq \overrightarrow{0}$, that also means $Ax \neq \lambda x $? $\endgroup$ – user1766555 Oct 19 '15 at 1:53
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    $\begingroup$ @user1766555 Specifically for the eigenvalue $\lambda=0$, yes. The only way for two vectors to be equal is if all entries are equal. The only way two real numbers multiplied together equals zero is if at least one of them is zero, but we said that $k\neq 0$ and $2\neq 0$, so $2k\neq 0$. As mentioned in my edit, there will be another eigenvector which does not have the property that the entries sum to zero, but this will be for a different eigenvalue. $\endgroup$ – JMoravitz Oct 19 '15 at 1:58
  • $\begingroup$ But it is not "without calculation". You must perform calculations to confirm that you are dealing with a multiple of the Ones matrix. Just because those calculations are done automatically in your head does not mean they do not occur. $\endgroup$ – mathreadler Oct 19 '15 at 2:10

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