Geometric multiplicity of eigenvalue 0 for 5×5 matrix of ones

[chuy52506]

Homework Statement

It is a 5x5 matrix with 1s in all of its entries.

Homework Equations

Find the geometric multiplicity of $$\lambda$$=0 as an eigenvalue of the matrix.

The Attempt at a Solution

WHat i did was use the characteristic equation of A-$$\lambda$$I
and then row reduce it. After that I am not sure what to do?

 

[lanedance]

how about taking the determinant?

the elementary row operations should not change the lambda^n term in the factored characteristic equation, only its multiplier

 

[HallsofIvy]

I'm not sure what lanedance meant about "taking the determinant". Obviously the determinant is 0. Perhaps he meant finding the characteristice equation which will obviously have $\lambda= 0$ as a multiple root but that would only tell you the algebraic multiplicity, not the geometric multiplicity (which can be any positive integer up to the algebraic multiplicity).

The geometric multiplicity of an eigenvalue is the number of independent eigenvectors corresponding to that eigenvalue, the dimension of the "eigenspace" correspondimng to that eigenvalue.

Here, you need to determine the number of independent solutions to
$$\begin{bmatrix}1 & 1 & 1 & 1 & 1 \\1 & 1 & 1 & 1 & 1\\1 & 1 & 1 & 1 & 1\\1 & 1 & 1 & 1 & 1\\1 & 1 & 1 & 1 & 1\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\\ 0 \\ 0\end{bmatrix}$$

That reduces to
$$\begin{bmatrix}x_1+ x_2+ x_3+ x_4+ x_5\\x_1+ x_2+ x_3+ x_4+ x_5\\x_1+ x_2+ x_3+ x_4+ x_5\\x_1+ x_2+ x_3+ x_4+ x_5\\x_1+ x_2+ x_3+ x_4+ x_5\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

or the single equation $x_1+ x_2+ x_3+ x_4+ x_5= 0$.

That is a single equation in 5 unknowns so you can solve for one of them in terms of the other four. You should be able to easily write down a basis for the "eigenspace" now.