20,790 views

Consider a matrix P whose only eigenvectors are the multiples of $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$.

Consider the following statements.

1. P does not have an inverse
2. P has a repeated eigenvalue
3. P cannot be diagonalized

Which one of the following options is correct?

1. Only I and III are necessarily true
2. Only II is necessarily true
3. Only I and II are necessarily true
4. Only II and III are necessarily true

edited
1. A necessary and sufficient condition for a matrix $A_{n\times n}$ to be diagonalisable is that the matrix must have $n$ linearly independent eigen vectors.
2. A sufficient condition for a matrix $A_{n\times n}$ to be diagonalisable is that the matrix must have $n$ linearly independent eigen values.

edited by
@kushagra     what is meaning of linearly independent eigen values?

And why 2nd statement is only sufficient condition?
@Gyanu I am not very sure of it. Maybe while I was going through vectors I misunderstood something and wrote those points. I felt that $2)$ one is not right. I Strike that for now but if got something I will update it.

Theorem: Suppose the $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors. If these eigenvectors are the columns of a matrix $S,$ then $S^{-1}AS$ is a diagonal matrix $\Lambda$. The eigenvalues of $A$ are on the diagonal of $\Lambda$.

$S^{-1}AS = \Lambda$ (A diagonal Matrix with diagonal values representing eigen values of A)$= \begin{bmatrix} \lambda _{1} & & & & \\ & \lambda _{2} & & & \\ & & . & & \\ & & & . & \lambda _{n}\\ & & & & \end{bmatrix}$

Now if $A$ is diagonalizable, $S^{-1}$ must exist. What is $S?$ $S$ is a matrix whose columns are eigen-vectors of matrix $A.$

Now $S^{-1}$ would only exist if $S$ is invertible. And $S$ would be invertible if all rows and columns of $S$ are independent.

If we would have same eigen-vectors, then $S^{-1}$ won't exist and hence $A$ won't be diagonalizable.

Even if a matrix $A$ has same eigen values, it does not mean that it is not diagonizable. Take the trivial example of the Identity Matrix $I.$

$\begin{bmatrix} 1 & & \\ & 1& \\ & & 1 \end{bmatrix}$

The Eigen values are $1,1,1$ and Eigen vectors are $\begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}$ , $\begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}$, $\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}$

We form $S=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 & 0& 1 \end{bmatrix}$ and this is invertible.

So, this makes $S^{-1} I S$ as $I =\Lambda$ which is in-fact a diagonal matrix.

So, even if we have same eigenvalues the matrix may or may not be diagonalizable. But yes, we need full $n$ set of linearly independent eigenvectors for this matrix $A$ of size $n \times n$ to be diagonalizable.

Now our problem says that we have a matrix $P$ whose only eigenvectors are multiples of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$.

Means we have only  $\begin{bmatrix} 1\\ 4 \end{bmatrix}$ as independent eigen vector. Surely, this matrix is not diagonalizable.

Since, eigen vectors are multiples of  $\begin{bmatrix} 1\\ 4 \end{bmatrix}$, we have repeated eigen values.

Let us assume $\lambda _{1}$ and $\lambda _{2}$ are different eigen values.

Since, we have only eigen vectors multiple of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$.

Let this vector be $x_1.$

So, $Px_1=\lambda _{1}x_1$

$\implies Px_1=\lambda _{2}x_1$

$\implies \lambda _{1}x_1=\lambda _{2}x_1$

Multiply by $x_1-1$ we get $\lambda _{1} =\lambda _{2}.$ Means, same Eigen vector gives same Eigen Value.

So yes, $P$ has repeated Eigen Values. (II) statement is true.

Now Statement (I) is not true. We cannot say this statement with exact surety.

Consider a matrix $A=\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$

It's eigenvales are $\lambda _{1}=\lambda _{2}=0$.

All the eigenvectors of this $A$ are multiples of vector $(1,0)$

$\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}. x=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

or $x=\begin{bmatrix} c\\ 0 \end{bmatrix}$.

This matrix surely is not diagonalizable but this matrix $A$ has determinant $= -1.$

Since determinant is not $0,$ so $A^{-1}$ exists.

Hence, (II) and (III) are surely valid under all cases of this question.

For (I) part, Matrix $P$ will not have an inverse when $\det(P)=0$ and this implies one of the eigenvalue of $P$ is zero. But in question, no where it is mentioned about what are the eigen values. So, I is not necessarily true.

Answer-(D) (Remember options says "necessarily true")

edited by

This matrix surely is not diagonalizable but this matrix A has determinant =−1.

## How determinant of A is -1???????

Please correct this part. Rest of the answer is very well explained. Thanks.

edited by

@Ayush Sir

how Identity matrix have eigen vectors [1,0,0] , [0,1,0] and [0,0,1] ?

I think Any vector is eligible to be a eigen vector of identity matrix.

since eigen value= 1,1,1

so AX= λX

(I-1.I)X=0    ….....……..(1)

equation (1) satify by any vector X.

correct me if I’m wrong.

A has Eigen value λ1 and λ2 which is  0 mention by you then in next line  u have written A has inverse only possible when determinant not equal to 0 but if atleast one Eigen value is 0 determinant is always 0 so by that A doesn't have inverse finally u are contradicting ur own statement only at the end.

What u want to say is for some matrix to have inverse none of the Eigen value can be 0 and here we don't know exactly what are Eigen values so we cannot say whether A has a inverse or not.

Ur example taken here is wrong otherwise all are fine.

Eigenvectors are multiple of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$ ,

So, it has repeated eigen value.

Hence, It cannot be Diagonalizable since repeated eigenvalue, [ we know if distinct eigen vector then diagonalizable]

$D$ must be answer.

If all eigenvalues are different, then all eigenvectors are linearly independent and all geometric and algebraic multiplicities are 1.

Say $\begin{bmatrix} 1\\ 4 \end{bmatrix}$ is multiple of $2$

that means eigen vectors are $2.\begin{bmatrix} 1\\ 4 \end{bmatrix}$=$\begin{bmatrix} 2\\ 8 \end{bmatrix}$

does it mean it has repeated eigen value?

Yes

It is given that the only eigen vector of the matrix P is [1 4]T, from this we can conclude that the matrix is atleast an 2x2 matrix, since the eigen vector given has 'm' components which we can take as 'm'  and since it is not possible to find eigen values and eigen vectors for rectangular matrices since they are not defined, minimum value of 'n' can be taken as 2.

Now we can eliminate option 1 : Because we can't say for sure whether the matrix has an inverse or not from the given eigen vector.

Option 2 : Since the minimum matrix possible is 2x2 and it is given that there is only one eigen vector, we can conclude that the matrix has repeated eigen values because, if the  matrix has distinct eigen values, then we would have got a characteristic equation in second degree polynomial with distinct roots and so there would have been 2 distinct eigen vectors. However we have only one eigen vector, So option 2 is correct

Option 3 : Generally, an (n x n) matrix with repeated eigen values can be diagonalised if we obtain n linearly independent eigen vectors for it. This will be the case; if for each repeated eigen value(i) of algebraic multiplicity(num. of times, the eigen value is repeated) N, we get 'N' linearly independent eigen vectors. So, in the options provided, it is given that "Option 3 is necessary true", we can conclude that option d) is correct. Because repeated eigen values are just necessary but not sufficient condition to say whether a matrix is diagonalisable or not.

Please note that there are some matrices which have repeated eigen values but still can be diagonalisable. So just because you have repeated eigen values, you can't conclude that the matrix can't be diagonalisable. So, the usage of the word "necessarily" in the option d) makes it the correct one. if "necessarily" has not been used, we can't conclude that option d) is the correct answer.

### 1 comment

here ....

even the eigen vectors also repeated ( geometric multiplicity is greater than 1 ).

For diagnolisation necessary and sufficient condition is to have n lineary independt eigen vectors.

But here we have only 1 linear independent eigen vector ( remaining are multiple of it ).

Hence diagnolisation is not possible.

There is no connection between diagonalizability (n independent eigenvectors) and invertibility (no zero eigenvalues).

The only indication given by the eigenvalues is: Diagonalization can fail only if there are repeated eigenvalues.

If all the eigenvectors are independent, then the matrix is diagonalizable. Here, it isn't the case, hence the matrix is not diagonalizable.

Since the only way the diagonalizability can fail is to have repeated eigenvalues (this is a necessary, but not a sufficient condition); hence our matrix has repeated eigenvalues.

Nothing can be said about invertibility, IMO. We need to know about eigenvalues to talk about it.

So, Option D

### A slightly different approach

If all eigenvalues are distinct => all eigenvectors are independent (equivalent to saying the matrix is diagonalizable) $[1]$

If the eigenvalues are repeated => eigenvectors may or may not be independent. Hence, we can't say if the matrix is diagonalizable or not.

But if we proved that the matrix is not diagonalizable, then it can't have distinct eigenvalues. Because $[1]$

So, can we prove that the matrix is not diagonalizable? Yes. Since eigenvectors are not independent, matrix is not diagonalizable.

So, Statements II and III are necessarily true.

If all the eigenvectors are independent, then the matrix is diagonalizable. Here, it isn't the case, hence the matrix is not diagonalizable.

$\text{Independent Eigen vectors } \rightarrow \text{Matrix is diagonalizable}$

So premise is false then how are you saying Matrix is not diagonalizable?
Iff*

So that will be: Independent Eigenvectors $\leftrightarrow$ Matrix is diagonalizable