The matrix A = ≤ft[ {*20c 3/2&0&1/2\\ 0& - 1&0\\ 1/2&0&3/2 } ] has three distinct Eigen values and one of its Eigen vectors is ≤ft[ {*20c 1\\ 0\\ 1 } ]. Which one of the following can be another Eigen vector of A?

Understanding Eigenvectors and Eigenvalues

An eigenvector of a square matrix is a non-zero vector that, when the matrix is multiplied by this vector, results in a scalar multiple of the same vector. This relationship is defined by the equation $Av=\lambda v$, where:

- $A$ is the square matrix.
- $v$ is the non-zero eigenvector.
- $\lambda$ is the corresponding eigenvalue (a scalar).

We are given the matrix $A$ and one of its eigenvectors. Our goal is to find which of the provided options is also an eigenvector of $A$.

Verifying the Given Eigenvector

First, let's confirm the properties of the provided eigenvector $v_1=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$ for the matrix:

$A=\left[\begin{array}{ccc}\frac{3}{2}& 0& \frac{1}{2}\\ 0& -1& 0\\ \frac{1}{2}& 0& \frac{3}{2}\end{array}\right]$

Calculate the product $A{v}_1$:

$A{v}_1=\left[\begin{array}{ccc}\frac{3}{2}& 0& \frac{1}{2}\\ 0& -1& 0\\ \frac{1}{2}& 0& \frac{3}{2}\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}(1)+0(0)+\frac{1}{2}(1)\\ 0(1)+(-1)(0)+0(1)\\ \frac{1}{2}(1)+0(0)+\frac{3}{2}(1)\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}+\frac{1}{2}\\ 0\\ \frac{1}{2}+\frac{3}{2}\end{array}\right]=\left[\begin{array}{c}2\\ 0\\ 2\end{array}\right]$

We observe that $A{v}_1=\left[\begin{array}{c}2\\ 0\\ 2\end{array}\right]$, which is equal to $2\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$. Thus, $A{v}_1=2v_1$. This confirms that $v_1$ is an eigenvector corresponding to the eigenvalue $\lambda =2$.

Testing Potential Eigenvectors

To find another eigenvector, we test each option by checking if it satisfies the eigenvector condition $Av=\lambda v$. We perform the matrix multiplication $Av$ for each potential vector $v$ and see if the result is a scalar multiple of $v$.

Option 1 Check

Consider the vector $v_{opt1}=\left[\begin{array}{c}0\\ 0\\ -1\end{array}\right]$.

Calculate $A{v}_{opt1}$:

$A{v}_{opt1}=\left[\begin{array}{ccc}\frac{3}{2}& 0& \frac{1}{2}\\ 0& -1& 0\\ \frac{1}{2}& 0& \frac{3}{2}\end{array}\right]\left[\begin{array}{c}0\\ 0\\ -1\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}(0)+0(0)+\frac{1}{2}(-1)\\ 0(0)+(-1)(0)+0(-1)\\ \frac{1}{2}(0)+0(0)+\frac{3}{2}(-1)\end{array}\right]=\left[\begin{array}{c}-\frac{1}{2}\\ 0\\ -\frac{3}{2}\end{array}\right]$

For $v_{opt1}$ to be an eigenvector, $A{v}_{opt1}$ must be $\lambda v_{opt1}$. Comparing $\left[\begin{array}{c}-\frac{1}{2}\\ 0\\ -\frac{3}{2}\end{array}\right]$ with $\lambda \left[\begin{array}{c}0\\ 0\\ -1\end{array}\right]$, we see that the first component $-1/2$ cannot equal $\lambda \times 0$. Thus, Option 1 is not an eigenvector.

Option 2 Check

Consider the vector $v_{opt2}=\left[\begin{array}{c}-1\\ 0\\ 0\end{array}\right]$.

Calculate $A{v}_{opt2}$:

$A{v}_{opt2}=\left[\begin{array}{ccc}\frac{3}{2}& 0& \frac{1}{2}\\ 0& -1& 0\\ \frac{1}{2}& 0& \frac{3}{2}\end{array}\right]\left[\begin{array}{c}-1\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}(-1)+0(0)+\frac{1}{2}(0)\\ 0(-1)+(-1)(0)+0(0)\\ \frac{1}{2}(-1)+0(0)+\frac{3}{2}(0)\end{array}\right]=\left[\begin{array}{c}-\frac{3}{2}\\ 0\\ -\frac{1}{2}\end{array}\right]$

Comparing $\left[\begin{array}{c}-\frac{3}{2}\\ 0\\ -\frac{1}{2}\end{array}\right]$ with $\lambda \left[\begin{array}{c}-1\\ 0\\ 0\end{array}\right]$, the third component $-1/2$ cannot equal $\lambda \times 0$. Thus, Option 2 is not an eigenvector.

Option 3 Check

Consider the vector $v_{opt3}=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$.

Calculate $A{v}_{opt3}$:

$A{v}_{opt3}=\left[\begin{array}{ccc}\frac{3}{2}& 0& \frac{1}{2}\\ 0& -1& 0\\ \frac{1}{2}& 0& \frac{3}{2}\end{array}\right]\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}(1)+0(0)+\frac{1}{2}(-1)\\ 0(1)+(-1)(0)+0(-1)\\ \frac{1}{2}(1)+0(0)+\frac{3}{2}(-1)\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}-\frac{1}{2}\\ 0\\ \frac{1}{2}-\frac{3}{2}\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$

We see that $A{v}_{opt3}=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$, which is equal to $1\times \left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$. This fits the definition $Av=\lambda v$ with $\lambda =1$. Thus, Option 3 is another eigenvector of $A$.

Option 4 Check

Consider the vector $v_{opt4}=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$.

Calculate $A{v}_{opt4}$:

$A{v}_{opt4}=\left[\begin{array}{ccc}\frac{3}{2}& 0& \frac{1}{2}\\ 0& -1& 0\\ \frac{1}{2}& 0& \frac{3}{2}\end{array}\right]\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}(1)+0(-1)+\frac{1}{2}(1)\\ 0(1)+(-1)(-1)+0(1)\\ \frac{1}{2}(1)+0(-1)+\frac{3}{2}(1)\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}+\frac{1}{2}\\ 1\\ \frac{1}{2}+\frac{3}{2}\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 2\end{array}\right]$

We check if $\left[\begin{array}{c}2\\ 1\\ 2\end{array}\right]=\lambda \left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$. From the first component, $2=\lambda (1)$, implying $\lambda =2$. However, for the second component, $1=\lambda (-1)$, implying $\lambda =-1$. Since we get different values for $\lambda$, Option 4 is not an eigenvector.

Final Conclusion

After checking all the options, only Option 3 satisfies the fundamental condition for an eigenvector, $Av=\lambda v$. Therefore, $\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$ is another eigenvector of the matrix $A$.