True, false questions about matrices

[lukaszh]

Hello,
could you help me with some true false questions about matrices?

First
If \mathbf{A}=\mathbf{B}-\mathrm{i}\mathbf{C} is hermitian matrix \mathbf{B},\mathbf{C} are real, then \mathbf{B},\mathbf{C} are anti-symmetric matrices. True? False?

My solution
If \mathbf{A} is hermitian, then \mathbf{A}^{\mathrm{H}}=\mathbf{A} so (\mathbf{B}-\mathrm{i}\mathbf{C})^{\mathrm{H}}=\mathbf{B}^{\mathrm{H}}-(\mathrm{i}\mathbf{C})^{\mathrm{H}}=\mathbf{B}^{\mathrm{T}}+i\mathbf{C}^{\mathrm{T}}. It implies the fact \mathbf{B}^{\mathrm{T}}+i\mathbf{C}^{\mathrm{T}}=\mathbf{B}-\mathrm{i}\mathbf{C}. \mathbf{B} is symmetric and \mathbf{C} is anti-symmetric. FALSE

Second
If A is diagonalizable and its eigenvalues are \{\lambda_1,\lambda_2,\cdots,\lambda_n\}, then \prod_{k=1}^{n}(x-\lambda_k)=0 has n different solutions

My solution
Matrix is diagonalizable, then \lambda_i\ne\lambda_j for i\ne j. So polynomial has n different soln's. TRUE


Thank you very much for your help...

 

[HallsofIvy]

You have a much bigger problem that whether this statement is true of false! What makes you think that if a matrix is diagonalizable, then \lambda_i\ne\lambda_j?

The identity matrix is diagonalizable because it is diagonal. What are its eigenvalues?

(An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. It doesn't matter what the eigenvalues are.)