2,002 views

​​​​​Which of the following is/are the eigenvector(s) for the matrix given below?

$$\begin{pmatrix} – 9 & – 6 & – 2 & – 4 \\ – 8 & – 6 & – 3 & – 1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix}$$

1. $\begin{pmatrix} – 1 \\ 1 \\ 0 \\ 1 \end{pmatrix}$
2. $\begin{pmatrix} 1 \\ 0 \\ – 1 \\ 0 \end{pmatrix}$
3. $\begin{pmatrix} – 1 \\ 0 \\ 2 \\ 2 \end{pmatrix}$
4. $\begin{pmatrix} 0 \\ 1 \\ – 3 \\ 0 \end{pmatrix}$

$Ax = \lambda x$
Learn very powerful and important way to multiply matrix and vector. $Ax$ is just linear combination of columns of $A$.. if you can understand this then solution of system of linear equation is just cakewalk for you. And complete linear algebra will come very intutive.

A.

$\Bigg[\begin{smallmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{smallmatrix} \Bigg] \Bigg[\begin{smallmatrix} -1\\ 1\\ 0 \\1 \end{smallmatrix} \Bigg] = -1 \Bigg[\begin{smallmatrix} -9\\ -8\\ 20 \\32 \end{smallmatrix} \Bigg] + 1 \Bigg[\begin{smallmatrix} -6\\ -6\\ 15 \\21 \end{smallmatrix} \Bigg] + 0 \Bigg[\begin{smallmatrix} -2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg] + 1 \Bigg[\begin{smallmatrix} -4\\ -1\\ 5 \\12 \end{smallmatrix} \Bigg] = \Bigg[\begin{smallmatrix} -1\\1 \\ 0 \\1 \end{smallmatrix} \Bigg]$

B.

$\Bigg[\begin{smallmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{smallmatrix} \Bigg] \Bigg[\begin{smallmatrix} 1\\ 0\\ -1 \\0 \end{smallmatrix} \Bigg] = 1 \Bigg[\begin{smallmatrix} -9\\ -8\\ 20 \\32 \end{smallmatrix} \Bigg] + 0 \Bigg[\begin{smallmatrix} -6\\ -6\\ 15 \\21 \end{smallmatrix} \Bigg] -1 \Bigg[\begin{smallmatrix} -2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg] + 0 \Bigg[\begin{smallmatrix} -4\\ -1\\ 5 \\12 \end{smallmatrix} \Bigg] = \Bigg[\begin{smallmatrix} -7\\-5 \\ 12 \\25 \end{smallmatrix} \Bigg]$

C.

$\Bigg[\begin{smallmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{smallmatrix} \Bigg] \Bigg[\begin{smallmatrix} -1\\ 0\\ 2 \\2 \end{smallmatrix} \Bigg] = -1 \Bigg[\begin{smallmatrix} -9\\ -8\\ 20 \\32 \end{smallmatrix} \Bigg] + 0 \Bigg[\begin{smallmatrix} -6\\ -6\\ 15 \\21 \end{smallmatrix} \Bigg] + 2 \Bigg[\begin{smallmatrix} -2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg] + 2 \Bigg[\begin{smallmatrix} -4\\ -1\\ 5 \\12 \end{smallmatrix} \Bigg] = 3\Bigg[\begin{smallmatrix} -1\\ 0\\ 2\\2 \end{smallmatrix} \Bigg]$

D.

$\Bigg[\begin{smallmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{smallmatrix} \Bigg] \Bigg[\begin{smallmatrix} 0\\ 1\\ -3 \\0 \end{smallmatrix} \Bigg] = 0 \Bigg[\begin{smallmatrix} -9\\ -8\\ 20 \\32 \end{smallmatrix} \Bigg] + 1 \Bigg[\begin{smallmatrix} -6\\ -6\\ 15 \\21 \end{smallmatrix} \Bigg] -3 \Bigg[\begin{smallmatrix} -2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg] + 0 \Bigg[\begin{smallmatrix} -4\\ -1\\ 5 \\12 \end{smallmatrix} \Bigg] = 3\Bigg[\begin{smallmatrix} 0\\ 1\\ -3 \\0 \end{smallmatrix} \Bigg]$

Doing it traditionally takes more time, at least happened to me during the exam! :(

Wish I knew this column picture before the exam, super-duper intuitive & cool
as in option C and D both eigen values are same but the eigen vectors are different ,as per rule every eigen value possess an unique eigen vector but how the both option c and d are correct simultaneously

@Nalluri.Sasikanth Good observation.

BUT, You are wrong in saying that “ every eigen value possess an unique eigen vector ”.

Two different eigenvalues has two different eigenvector  -TRUE

But the  eigen value may also have different eigenvectors.

I have given all PROOFS in the video. You will definitely like this.

Same eigen values can may or may not have same eigen vectors.

For different eigen values we will definitely have different eigen vectors