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I am having difficulties to understand how can I calculate the Geometric Multiplicity of this matrix:

In this case I have the matrix:

A= \begin{bmatrix} 1 & 0 \\ 0 & 9 \end{bmatrix} \quad

where I obtained the matrix:

B= \begin{bmatrix} 1-\lambda & 0 \\ 0 & 9-\lambda \end{bmatrix}

and by finding the characteristic equation I obtained the eigenvalues which are:

$\lambda=5-\sqrt(34)$ and $\lambda=5+\sqrt(34)$

So, the Algebraic Multiplicity is 1 ( no repeated roots).

Then to obtain the Geometric Multiplicity I know that I need to find:

B= \begin{bmatrix} 1-\lambda & 0 \\ 0 & 9-\lambda \end{bmatrix} and substitute $\lambda=5-\sqrt(34)$ and $\lambda=5+\sqrt(34)$

so I got:

\begin{bmatrix} 1-(5-\sqrt(34)) & 0 \\ 0 & 9-(5-\sqrt(34)) \end{bmatrix}

= \begin{bmatrix} -4+\sqrt(34)) & 0 \\ 0 & 4+\sqrt(34)) \end{bmatrix}

By applying elementary row operations I obtained the matrix

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} and could conclude that the rank is 2 and dimension of the null space ( dimension of kernel) is zero. So I think that the geometric multiplicity is zero. However we have the property that: Geometric multiplicity is:

1<=Geometric multiplicity<=Algebraic Multiplicity and in this case: 1<=0<=1 does not make any sense.

Can anyone help me on this?

Thanks

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    The geometric multiplicity is never $0$. It is the maximal possible number of linear independent eigenvectors to a given eigenvalue. – 2017-02-13

1 Answers 1

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The characteristic equation is actually $(1-\lambda)(9-\lambda)=0$ so the eigenvalues are $1$ and $9$. You made a mistake calculating the characteristic polynomial.

However, this matrix does not require this kind of analysis. This is one of a number of situations where the eigenvalues and eigenvectors are obvious. See this lesson http://lem.ma/C on Lemma for the diagonal-matrix "giveaway" and the subsequent lessons for other easy giveaways.

In particular, the eigenvalues of this matrix are $1$ and $9$ and the corresponding eigenvectors are $$\begin{bmatrix}1\\0\end{bmatrix} \text{ and } \begin{bmatrix}0\\1\end{bmatrix}.$$

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    You are right I did a mistake in the calculation thank you for your help. However I do not understand how can I find the Geometric multiplicity. Do I need to know the Dimension of the null space? thanks – 2017-02-13
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    You don't need to know it. But in case you are curious, the null space consists of the zero vector alone (since the vectors are linearly independent) and is therefore zero-dimensional. – 2017-02-14
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    I thought that to find the Geometric multiplicity we needed to find the dimension of the null space and then in this case as you said the null space is the zero vector and the dimension of the null space is 1 and consequently the G.M is also 1 – 2017-02-14