 # 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 = 4 is two-dimensional. $A=\begin{bmatrix} 4&2&3&3 \\ 0&2 &h&3 \\ 0&0&4&14 \\ 0&0&0&2\end{bmatrix}$

This problem aims to familiarize us with eigenvalues, eigenspace, and echelon form. The concepts required to solve this problem are related to basic matrices which include eigenvectors, eigenspace, and row reduce forms.

Now, eigenvalues are a unique set of scalar numbers that are linked with the linear equations which can be found in the matrix equations. Whereas the eigenvectors, also known as characteristic roots, are basically non-zero vectors that can be altered by their scalar element when of course linear transformation is applied.

In the statement, we are given the eigenspace which is basically the set of eigenvectors linked with each eigenvalue when the linear transformation is applied to those eigenvectors. If we recall linear transformation, it is often in the form of a square matrix whose columns and rows are of the same count.

To find out the value of $h$ for which the $\lambda = 4$ is two-dimensional, we first have to convert the matrix $A$ to its echelon form.

Firstly performing the operation $A- \lambda I$, where $\Lambda = 4$ and $I$ is the identity matrix.

$A = \begin{bmatrix} 4&2&3&3 \\ 0&2&h&3 \\ 0&0&4&14 \\ 0&0&0&2\end{bmatrix} – 4 \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

$= \begin{bmatrix} 4&2&3&3 \\ 0&2&h&3 \\ 0&0&4&14 \\ 0&0&0&2\end{bmatrix} – \begin{bmatrix} 4&0&0&0 \\ 0&4&0&0 \\ 0&0&4&0 \\ 0&0&0&4 \end{bmatrix}$

$A = \begin{bmatrix} 0&2&3&3 \\ 0&-2&h&3 \\ 0&0&0&14 \\ 0&0&0&-2\end{bmatrix}$

To make $0$ on second pivot, applying the operation $R_2 \rightarrow R_2 + R_1$, Matrix $A$ becomes:

$A = \begin{bmatrix} 0&2&3&3 \\ 0&0&h+3&6 \\ 0&0&0&14 \\ 0&0&0 &-2\end{bmatrix}$

Now dividing $R_3$ with the $14$ and performing the operation $R_4 \rightarrow R_4 – R_3$, Matrix $A$ becomes:

$A = \begin{bmatrix} 0& 2& 3& 3 \\ 0& 0& h+3& 6 \\ 0& 0& 0& 1 \\ 0& 0& 0& 0 \end{bmatrix}$

By Looking at the echelon form of the matrix $A$, it can be deduced that variable $x_1$ is a free variable if $h \neq -3$.

If $h= -3$, then it is not in echelon form, but the only one-row operation is needed it into echelon form. In that case, $x_1$ and $x_2$ will be the free variable so the eigenspace it produces will be two-dimensional.

## Numerical Result

For $h = -3$ the eigenspace of $\lambda = 4$ is two-dimensional.

## Example

Find $h$ in the matrix $A$ such that the eigenspace for $\lambda = 5$ is two-dimensional.

$A = \begin{bmatrix} 5 &-2 &6 &-1 \\ 0 &3 &h &0 \\ 0 &0 &5 &4 \\ 0 &0& 0& 1 \end{bmatrix}$

The echelon form of this matrix can be obtained by applying some operations and it comes out to be:

$A = \begin{bmatrix} 0& 1& -3& 0 \\ 0 &0 &h-6 &0 \\ 0 &0 &0 &1 \\ 0 &0 &0 &0 \end{bmatrix}$

It can be seen for $h =6$ the system will have $2$ free variables and hence it will have an eigenspace of two-dimensional.