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Q5.2-18E

Expert-verified
Found in: Page 267

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Question 18: It can be shown that the algebraic multiplicity of an eigenvalue $$\lambda$$ is always greater than or equal to the dimension of the eigenspace corresponding to $$\lambda$$. Find $$h$$ in the matrix $$A$$ below such that the eigenspace for $$\lambda = 5$$ is two-dimensional:$A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]$

It is observed that the system requires two free variables for a two-dimensional eigenspace. This occurs only when $$h = 6$$.

See the step by step solution

## Step 1: Definition of the characteristic polynomial

The eigenvalue of an $$n \times n$$ matrix $$A$$ is a scalar $$\lambda$$ such that $$\lambda$$ satisfies the characteristic equation $$\det \left( {A - \lambda I} \right) = 0$$.

When $$A$$ is a $$n \times n$$ matrix, $$\det \left( {A - \lambda I} \right)$$ is the characteristic polynomial of $$A$$ which is the polynomial of degree $$n$$.

In particular, the multiplicity of an eigenvalue $$\lambda$$ represents its multiplication as a root of the characteristic equation.

## Step 2: Determine h in matrix A

Write the matrix in the $\left( {A - 5I} \right)$ form, as shown below.

$\begin{array}{c}\left( {{\rm{A}} - 5I} \right) = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&{\rm{h}}&0\\0&0&5&4\\0&0&0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}5&0&0&0\\0&5&0&0\\0&0&5&0\\0&0&0&5\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&{ - 2}&{\rm{h}}&0\\0&0&0&4\\0&0&0&{ - 4}\end{array}} \right]\end{array}$

Write the matrix as an augmented matrix $$\left( {A - 5I} \right)x = 0$$, as shown below.

$$\sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&{ - 2}&{\rm{h}}&0&0\\0&0&0&4&0\\0&0&0&{ - 4}&0\end{array}} \right]$$

Apply row operation in the augmented matrix:

At row 2, subtract row 1 from row 2. At row 4, multiply row 4 by $$- 1$$, as shown below.

$$\sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&4&0\\0&0&0&4&0\end{array}} \right]$$

At row 4, multiply row 3 by 1 and subtract it from row 4, as shown below.

$$\sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&4&0\\0&0&0&0&0\end{array}} \right]$$

At row 1, multiply row 1 by $$- \frac{1}{2}$$ and at row 3, multiply row 3 by $$\frac{1}{4}$$, as shown below.

$$\sim \left[ {\begin{array}{*{20}{c}}0&1&{ - 3}&1&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]$$

At row 1, subtract row 3 from row 1, and at row 2, subtract row 3 from row 2, as shown below.

$$\sim \left[ {\begin{array}{*{20}{c}}0&1&{ - 3}&0&0\\0&0&{{\rm{h}} - 6}&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]$$

The above system requires two free variables for a two-dimensional eigenspace. This occurs only when $$h = 6$$.

Thus, the value of $$h$$ in the matrix is 6.