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so I have the matrix:

$$A = \begin{bmatrix}1 & 0 & 0 & 0\\1 & 1 & 0 & 0 \\1 & 2 & 1 & 0 \\ 1 & 3 & 3 & 1 \end{bmatrix}$$

Since it's a triangular matrix, I know the eigenvalues are $\lambda_1=\lambda_2=\lambda_3=\lambda_4=1$, so $\lambda$ has an algebraic multiplicity of $4$. Once you row reduce everything to obtain the eigenvectors, I got:

$$x_1=0$$ $$x_2=0$$ $$x_3=0$$ $$x_4=t$$

So my question is, what is the geometric multiplicity of $\lambda$? Is it just $1$?

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  • $\begingroup$ Is $A_{4,4} = 0$ or $1$? $\endgroup$ – JimmyK4542 Jan 12 '15 at 4:37
  • $\begingroup$ @JimmyK4542 i messed up writing the matrix $\endgroup$ – Sofia June Jan 12 '15 at 4:39
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Since the geometric multiplicity is the dimension of the eigenspace, you have that geometric multiplicity of the eigenvalue $1$ is $1$, as there is only one linearly independent eigenvector.

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