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+16 votes

In questions $1.1$ to $1.7$ below, one or more of the alternatives are correct. Write the code letter$(s)$ $a$, $b$, $c$, $d$ corresponding to the correct alternative$(s) $ in the answer book.  Marks will be given only if all the correct alternatives have been selected and no incorrect alternative is picked up.

1.1). The eigen vector $(s)$ of the matrix

$$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$$ is (are)

  1. $(0,0,\alpha)$
  2. $(\alpha,0,0)$
  3. $(0,0,1)$
  4. $(0,\alpha,0)$
asked in Linear Algebra by Veteran (59.8k points)
edited by | 1.8k views
I believe the answer should be only B. It is because if we look at the matrix it shows the transformed 3D space into a line where the z^ vector is transformed into a line replicating x axis. In that scenario the only vector that remains unchanged is (alpha, 0, 0) i.e option B.
could anyone give the complete procedure ?
Answer is b only why obtion d is consider here ?
Why d is included in answer?

Here we got eign values as 0,0,0.

Now to get eign vector take  λ=0 in

( A-λI) X =0 or AX=0 -----(i)

here X is eign vector in form of column matrix [X1,X2,X3].

If we multiply as per (i) we get equation as

X1*0+X2*0+ aX3=0

Now for LHS to be 0 , X3 must be 0 . Here X1 and X2 can be assumed any value . As per the given option a and c are wrong.hence b and d satisfy the case
what about (alpha,alpha,0) vectors? will they be eigen vectors too.I think they are.
Bind blowing explanation satyajeet sir
Yes they will be eigen [email protected] gow

3 Answers

+12 votes
Best answer
Since, the given matrix is an upper triangular one, all eigenvalues are 0. And hence $A-\lambda I=A$

So the question asks

$\begin{bmatrix} 0 &0 &\alpha \\ 0&0 &0 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

what $x_1,x_2,x_3$ are suitable?

which means $x_1$times column1+$x_2$times column2+$x_3$ times column3=Zero Vector

Look carefully,if you take any linear combination of columns 1 and 2 and don't take column 3(means $x_3=0$), the result will be a zero vector.

So, if we have our X=$\begin{bmatrix} \beta\\ \gamma\\ 0 \end{bmatrix}$ where $\beta, \gamma$ are any scalar quantity,

but $x_3$ must be necessarily zero to get zero vector.

Hence, only option (B) and (D) satisfy.
answered by Boss (25.4k points)
selected by
The second part of the answer is difficult for me to grasp.
Which one?
From the statement "look carefully"
ohh that one. See from this example.

Suppose you have a system like

$\begin{bmatrix} 1 &3 &6 \\ 7 & 5& 9\\ 3 &8 & 2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$

this is equivalent to

$x_1\begin{bmatrix} 1\\ 7\\ 3 \end{bmatrix}+x_2\begin{bmatrix} 3\\ 5\\ 8 \end{bmatrix}+x_3\begin{bmatrix} 6\\ 9\\ 2 \end{bmatrix}$

In the question given, you have to produce a zero vector as result. What suitable combinations of $x_1,x_2,x_3$ can you take?
+15 votes
Answer: $B, D$

Eigen values are: $0,0,0$

The eigen vector should satisfy the equation: $\alpha$z $=$ $0$
answered by Boss (33.8k points)
edited by
yes, I also think ans should be option B only.
how b and d?
here by normal approach we get in eigen vector x3=0

So, answer will be b) and d)
^ Will $-\lambda^3 = 0$ not be the characterastic equations for this matrix?
So how third value came $\alpha$?
yes (B) and (D) should be correct option!

@Manu Thakur

here we have the Basic variable as "z" and free variable as "x" and "y".

So, keeping x = t and y = s, and z = $\alpha$

Then vector corresponding to x is $[1, 0, 0]^T$

And vector corresponding to y is $[0, 1, 0]^T$

Then why it is not $[1, 0, 0]^T$ And $[0, 1, 0]^T$ instead of $[\alpha, 0 , 0]^T$ And $[0, \alpha, 0]^T$?

$\alpha$ or 1 doesnot matter

atleast u have to give answer, from options given
But it should be $[1,0,0]^T$ And $[0,1,0]^T$ even no option matches. ryt??
if B & D are true then A should also be true please check once?
+4 votes
Given matrix is an upper triangular matrix. Therefore diagonal elements of A are eigen values of A, i.e. λ=0,0,0.

Consider $(A-λI)x=0$

$\Rightarrow$ $\begin{bmatrix} 0-\lambda & 0 & \alpha \\ 0 & 0-\lambda & 0\\ 0 & 0 & 0-\lambda \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Put λ=0

i.e. $\begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Now we have


$\Rightarrow$$x_3$ = 0

Therefore, any non zero vector with z-component as zero is an eigen vector.

Hence in the given options $B$ and $D$ are correct.
answered by Active (4.8k points)

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