The geometric mutiplicity of a matrix

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SUMMARY

The geometric multiplicity of the eigenvalue λ=0 for the given 5x5 matrix, where all entries are 1, is determined to be 4. This conclusion is reached by solving the equation represented by the matrix and analyzing its eigenvalues and eigenspaces using SageMath. The eigenvalues of the matrix are [5, 0, 0, 0, 0], indicating that the eigenspace corresponding to λ=0 has a dimension of 4, which confirms the geometric multiplicity.

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chuy52506
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It is a 5x5 matrix with 1s in all of its entries.
Find the geometric multiplicity of [tex]\lambda[/tex]=0 as an eigenvalue of the matrix.
 
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What have you done yourself?

Solve the equation
[tex]\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}[/tex].

How many distinct equations do you get relating those variables? Can you solve for some in terms of the others?
 
Here's the text of a http://www.sagenb.org/" that answers your question.
Since you've shown no work, I'll leave the interpretation up to you.

Code: 
sage: A = Matrix(QQ,[[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]]); A
[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]

sage: A.eigenvalues()
[5, 0, 0, 0, 0]

sage: A.eigenspaces()
[
(5, Vector space of degree 5 and dimension 1 over Rational Field
User basis matrix:
[1 1 1 1 1]),
(0, Vector space of degree 5 and dimension 4 over Rational Field
User basis matrix:
[ 1  0  0  0 -1]
[ 0  1  0  0 -1]
[ 0  0  1  0 -1]
[ 0  0  0  1 -1])
]
 
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