TIFR CSE 2019 | Part A | Question: 3

Question

$A$ is $n \times n$ square matrix for which the entries in every row sum to $1$. Consider the following statements:

1. The column vector $[1,1,\ldots,1]^T$ is an eigen vector of $A.$
2. $ \text{det}(A-I) = 0.$
3. $\text{det}(A) = 0.$

Which of the above statements must be TRUE?

• A. Only $(i)$
• B. Only $(ii)$
• C. Only $(i)$ and $(ii)$
• D. Only $(i)$ and $(iii)$
• E. $(i),(ii) \text{ and }(iii)$

Comments

if linearly independent or not eigen vector always change

It is case dependent

right?

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Linearly Independent vector$:-$     $rank(A)$ $<$ $Number$ $of$ $unknown(variables)$

So, in this case, we can assign  $" UK-rank(A)"$ values linearly independently to the $unknown(UK).$

Let's say $A=\begin{bmatrix} 1& 0\\0 &1 \end{bmatrix}_{2\times 2}$

        For Eigenvector       $AX=\lambda X$

                                         $AX-\lambda X=[0]$

                                         $(A-\lambda I )X=[0]$-------->$(1)$

We get Eigen values $\lambda_{1},\lambda_{2}=1$

                       Now,$\begin{bmatrix} 0& 0\\0 &0 \end{bmatrix}.\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

Here  $rank(A-\lambda I)=0$ and Number of variables(Unknown)$=2$

$rank(A)<UK\Rightarrow 0<2$

So, in this case, we can assign  $" UK-rank(A)"$ values linearly independently to the $unknown(UK).$

 $UK-rank(A-\lambda I)=2-0=2$

So,we can assign $x_{1}=k_{1}$ and $x_{2}=k_{2}$

$X=\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}=\begin{bmatrix} k_{1}\\ k_{2} \end{bmatrix}$

We can put $k_{1}=1$ and $k_{2}=1$

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check https://gateoverflow.in/31516/tifr-2015-maths-a-1

Answer 1

$(iii)$ is clealy $\text{incorrect}$. Example :$\begin{bmatrix} 1 &0 &0 \\ 0&0 &1 \\ 0&1 &0 \end{bmatrix}$

Determinant of this matrix is $-1$

Let the matrix $A=$ $\begin{bmatrix} x_{11} &x_{12} &x_{13} &\ldots \\ x_{21} &x_{22} &x_{23} &\ldots \\ x_{31} &x_{32} &x_{33} &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix}$ of order $n*n$

$\sum_{j=1}^{n}x_{ij}=1\,\,\forall\,\,i=1,\ldots,n$

We know, $Av = \lambda v$, where $\lambda$ are the eigenvalues and $v$ is the eigenvector

As,  $Av=v$

$\begin{bmatrix} x_{11} &x_{12} &x_{13} &\ldots \\ x_{21} &x_{22} &x_{23} &\ldots \\ x_{31} &x_{32} &x_{33} &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1\\ ...\\ \end{bmatrix} = \begin{bmatrix} x_{11}+x_{12}+x_{13}+\ldots\\ x_{21}+x_{22}+x_{23}+\ldots\\ x_{31}+x_{32}+x_{33}+\ldots\\ x_{41}+x_{42}+x_{43}+\ldots\\ \end{bmatrix}=\begin{bmatrix} 1\\ 1\\ 1\\ \ldots \\ \end{bmatrix}=v$

$v$ is an eigen vector and eigen value is $1$

So, $(i)$ is $correct$

$A\,-\,\lambda I = \begin{bmatrix} x_{11}-1 &x_{12} &x_{13} &\ldots \\ x_{21} &x_{22}-1 &x_{23} &\ldots \\ x_{31} &x_{32} &x_{33}-1 &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix}$

Now, $\sum_{j=1}^{n}x_{ij}=0\,\,\forall\,\,i=1,\ldots,n$

Applying Matrix transformation,

$C_{1} \leftarrow\sum_{j=1}^{n}C_{j}$

The matrix now becomes = $ \begin{bmatrix} 0 &x_{12} &x_{13} &\ldots \\ 0 &x_{22}-1 &x_{23} &\ldots \\ 0 &x_{32} &x_{33}-1 &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix}$

As $C_{1}$ is completely 0 , so $|A-\lambda I|=0$

$(ii)$ is $correct$

So, option (C).

Comments

the example you used for ($iii$) is counter example for ($i$)

https://www.emathhelp.net/calculators/linear-algebra/eigenvalue-and-eigenvector-calculator/?i=%5B%5B1%2C0%2C0%5D%2C%5B0%2C0%2C1%5D%2C%5B0%2C1%2C0%5D%5D&steps=on

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@Mk Utkarsh

what is problem with C)?

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i misunderstood C is correct

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$\begin{bmatrix} 1\\ 1 \end{bmatrix}$  can be

right?

not always $\begin{bmatrix} 0\\ 0 \end{bmatrix}$

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yes right

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@Mk Utkarsh

how can be sure that every eigen value is 1 here?

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himgta i never said that

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@himgta corrected it 

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@Shobhit Joshi

I m not getting how 1st statement is correct...for eigen value=1 it is getting true...but what about other cases?

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@himgta $\lambda=1$ proves that $v$ is a eigen vector, why check other values ?

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got it..thanks!

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Above answer explanation is good but i want to add some images that contains same answer but with some more information which makes solution, easy to understand. I don't know why uploaded images are not clear but if you want to understand the solution, please read it.

                           

Answer 2

Summation of all rows is $1$ implies

$AX=B$, where $A$ is given matrix and $B = [1,1,1,...]^T$. 

1. Putting $X=[1,1,1...]$ follows $AX=1.X$ implies $X$ is eigen vector with eigen value $(\lambda)$= $1$.
2. $|A-\lambda{I}|X = 0$ with $X \neq 0$ implies $|A-\lambda{I}|$ is singular.
3. $I_{2x2}$ is one of the counterexample. i.e, nothing can be said about $A$. It can either be singular or non singular.