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How many of the following matrices have an eigenvalue 1?
$\left[\begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right]\left[\begin{array}{cc}0 & 1 \\0 & 0 \end{array} \right] \left[\begin{array}{cc}1 & -1 \\1 & 1 \end{array} \right]$ and $\left[\begin{array}{cc}-1 & 0 \\1 & -1 \end{array} \right]$

1. one
2. two
3. three
4. four
0
In 3rd matrix. ..if we apply R2 -> R2-R1 then it transforms into upper triangular matrix and so getting eigen values 1 and 2 ...where am I wrong here.??
0
3rd matrix where  a11= 1 ,a12 = -1,a21=1,a22=1.

After R2->R2-R1

a21-a11=1-1=0

a22-a12=1-(-1)=2

SO it is upper triangular..

eigen values are diagonal entries..

So 1 and 2 are eigen values...isn't it.??
0
I think 1 & 2 will be the eigen values of $\begin{bmatrix} 1 & -1\\ 0& 2 \end{bmatrix}$ and not $\begin{bmatrix} 1 & -1\\ 1& 1 \end{bmatrix}$.
+5
If you add a row of A to another row, or exchange rows, the eigenvalues
usually change. Elimination does not preserve the Eigen values.

A better way is to check if | A- λI| = 0,  If 1 (given in the question) is the eigen value of this matrix. and the charactersitic equation satisfies. By doing so for all the matrix we find the equation satisfies for first  matrix.

Hence the answer is one , ie, option A

I found this way much faster then finding the values and then checking if the eigen values.
+2

@Ravi_1511

Elementary row operations change eigenvalues of given matrix (https://socratic.org/questions/do-elementary-row-operations-change-eigenvalues)

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Yes, you are right. It is valid in the case of finding the rank.

Characteristic equation is $| A- \lambda I | = 0$
(1)$$\begin{vmatrix}1- \lambda & 0\\ 0 & - \lambda\\ \end{vmatrix} = 0$$$(1- \lambda )(- \lambda) = 0$
$\implies \lambda =0,1$

Similarly, (2)  $\lambda =0,0$

(3) $\lambda \neq 1$

(4) $\lambda =-1,-1$

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@Keith, @Arjun, should not be for Matrix (4) eigenvalue = -1,-1 ? Please check. I think 0,-2 is not correct. IT does not change the final answer, someone crosscheck.  λ

+1

yUP IT WILL BE 0 AND -2.

equation become (-1-λ)2 -1= λ+2λ=0 now λ=0 and λ=-2

+1
From where you are getting that -1 . It should be (-1-lambda)^2 - 0.

When Last I saw, 1 * 0 = 0 !
+1

sory friend you are correct

(-1-λ)2

λ+2λ+1=0

λ= -1 ,-1

0
yes last one is -1,-1

just check det=1

eigen val cannot be 0,-2 coz then prod (eig)!=det(matrix)
+2
A bit faster method is to for each matrix, remove 1 from the diagonal elements, and then see if the determinant of the matrix is 0.This happens for only one matrix. And yes, this will work because $|A-\lambda I|=0$
0
If Eigen values are $\lambda_{1},\lambda_{2}$

We can write like this $x^{2}-(\lambda_{1}+\lambda_{2})x+\lambda_{1}.\lambda_{2}=0$

Out of given answers, 3 of them are (upper/lower) triangular matrices. So, that will give eigen values directly.

Property: The eigen value of triangular / unity / scalar / diagonal matrices is equal to principal diagonal elements itself.

3rd option need to be solved which wont be equal to 1.

So first matrix only have eigen value 1.

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