Give an example of a 2 times 2 matrix B which is diagonalizable but limn rightarrow infinity Bn does not exist? (Can you choose all eigenvalues with | lambda | 1?)
Solution
Consider the matrix given by: row 1 =[ 1 1 ], row 2 = [ 0 0 ] This is not invertible since the det = 0; or you can see it since the diagonal entries are the eigenvalues and one of them is zero, so it\'s not invertible. This matrix can be diagonalized. You know it by the following argument: This is a triangular matrix so the diagonal entries are the eigenvalues. So we see that this matrix has 2 distinct eigenvalues. Every eigenvalue must have an eigenvector. The eigenvectors of distinct eigenvalues are linearly independent, so this matrix has 2 genuine independent eigenvectors. Thus it is diagonalizable. If you didn\'t see this, you could work out the eigenvectors and see that in fact you get 2 independent ones. So it can be diagonalized.
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Give an example of a 2 times 2 matrix B which is diagonalizable but li.docx
1. Give an example of a 2 times 2 matrix B which is diagonalizable but limn rightarrow infinity Bn
does not exist? (Can you choose all eigenvalues with | lambda | 1?)
Solution
Consider the matrix given by: row 1 =[ 1 1 ], row 2 = [ 0 0 ] This is not invertible since the det =
0; or you can see it since the diagonal entries are the eigenvalues and one of them is zero, so it's
not invertible. This matrix can be diagonalized. You know it by the following argument: This is a
triangular matrix so the diagonal entries are the eigenvalues. So we see that this matrix has 2
distinct eigenvalues. Every eigenvalue must have an eigenvector. The eigenvectors of distinct
eigenvalues are linearly independent, so this matrix has 2 genuine independent eigenvectors.
Thus it is diagonalizable. If you didn't see this, you could work out the eigenvectors and see that
in fact you get 2 independent ones. So it can be diagonalized.
![Give an example of a 2 times 2 matrix B which is diagonalizable but limn rightarrow infinity Bn
does not exist? (Can you choose all eigenvalues with | lambda | 1?)
Solution
Consider the matrix given by: row 1 =[ 1 1 ], row 2 = [ 0 0 ] This is not invertible since the det =
0; or you can see it since the diagonal entries are the eigenvalues and one of them is zero, so it's
not invertible. This matrix can be diagonalized. You know it by the following argument: This is a
triangular matrix so the diagonal entries are the eigenvalues. So we see that this matrix has 2
distinct eigenvalues. Every eigenvalue must have an eigenvector. The eigenvectors of distinct
eigenvalues are linearly independent, so this matrix has 2 genuine independent eigenvectors.
Thus it is diagonalizable. If you didn't see this, you could work out the eigenvectors and see that
in fact you get 2 independent ones. So it can be diagonalized.](https://image.slidesharecdn.com/giveanexampleofa2times2matrixbwhichisdiagonalizablebutli-230127182543-641a8393/85/give-an-example-of-a-2-times-2-matrix-b-which-is-diagonalizable-but-lidocx-1-320.jpg)
