Block Diagonal Matrices

Every operator on a finite-dimensional complex vector space has a matrix (with respect to some basis of the vector space) that is a block diagonal matrix, with each block itself an upper-triangular matrix that contains only one eigenvalue on the diagonal.

[toddmorrill]

I really appreciate these videos and the context you provide when going through the results. But 8.21 and 8.29 were trickier proofs and I wish you wouldn't bail on these harder proofs where it would actually be helpful for you go through the results.

[raokondaMathematics]

How to find the eigenvalues of the matrix of type [ 0 A; B 0] . That is first row of the matrix is [0 A], and second row of the matrix is [B, 0], where A, B are square matrices of same order and 0 is a null matrix. Here order of Null matrix=order of A=Order of B

[maxgomez]

Hello,

Looking at the proof of 8.29 and considering the proof of 8.19 I am getting to the conclusion that the form promised by 8.29 will always be achieved if we consider a basis of generalized eigenvectors of the linear operator on V. This concern arised to me when looking at the end of example 8.30. I will appreciate any feedback.

[johnwitha]

I don't understand how the matrix of T with respect to basis (1, 0, 0), (0, 1, 0), (10, 2, 1) has a third column consisting of [0, 0, 7]. Would love some help.

[vipinshukla]

Thanks sir for such a great it would be helpful if you can clarify how (0, 1, 0) is an eigenspace corresponding to eigen value 6 @ 2:26 because I'm getting only the first one i.e., (1, 0, 0) as eigen vector corresponding 6.

[consideration]

I feel so confused why when the eigenvalue = 6 -> the eigenspace spans ((1, 0, 0), (0, 1, 0))? When I plug in lambda = 6 back to (T - lambda*1) = 0 and solve, I get z2 = z3 = 0, and z1 = any number. So, the eigensapce should span only (1, 0, 0).

[deepaksahu]

Can anyone provide me Example of 4×6 block lower triangular matrix ?
Thanks