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GO Classes DPP | GATE CS | Linear Algebra | Eigen Vectors

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Suppose that a $4 \times 4$ matrix $A$ has eigenvalue $3$. If

$A - 3I = \begin{pmatrix} 1 & -2 & 0 & 4 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix},$

then the eigenvectors of $A$ corresponding to the eigenvalue $3$ are of the form:

  1. $\begin{pmatrix} 2s - 4t \\ s \\ 3t \\ t \end{pmatrix} ; s,t \in \mathbb{R}$

     
  2. $\begin{pmatrix} 2s + 4t \\ s \\ 3t \\ t \end{pmatrix} ; s,t \in \mathbb{R}$

     
  3. $\begin{pmatrix} 2s - 4t \\ s \\ t \\ 3t \end{pmatrix} ; s,t \in \mathbb{R}$

     
  4. $\begin{pmatrix} 2s - 4t \\ 3t \\ s \\ t \end{pmatrix} ; s,t \in \mathbb{R}$

1 Answer

2 2 votes

If $3$ is an eigenvalue of $A$, then the corresponding eigenvectors satisfy

$(A-3I)\mathbf{x} = 0$.

Let $\mathbf{x} = \begin{pmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{pmatrix}$. Then from

$A - 3I = \begin{pmatrix} 1 & -2 & 0 & 4 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

we get the system

$x_1 - 2x_2 + 4x_4 = 0$

and

$x_3 - 3x_4 = 0$.

Since $x_2$ and $x_4$ are free variables, let

$x_2 = s$ and $x_4 = t$.

Then from $x_3 - 3x_4 = 0$, we get

$x_3 = 3t$.

From $x_1 - 2x_2 + 4x_4 = 0$, we get

$x_1 = 2s - 4t$.

So the eigenvectors are all vectors of the form $\begin{pmatrix} 2s - 4t \\ s \\ 3t \\ t \end{pmatrix} ; s,t \in \mathbb{R}$.

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