\[ A=\begin{bmatrix} 4&2&3&3 \\ 0&2 &h&3 \\ 0&0&4&14 \\ 0&0&0&2\end{bmatrix} \]

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.

## Expert Answer

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.**