You can use the definition of matrix powers at 12:36 in the video. First, diagonalize A as A = PDP^-1. Then raise each diagonal entry of D to the power of 2^(1/2), giving you the matrix D^(2^1/2). Then multiply to get A^(2^1/2) = P*D^(2^1/2)*P^-1.
Hi, is there a quick argument wich shows, that the definition given at around 13:00 for the power A^r of a matrix is well defined (does not depend on the chosen transformation matrix, i.e. the order of the eigenvalues)? Does this definition match with the definition A^r=exp(r log A), for the matrix logarithm / exponential? Thanks in advance :)
Good question! I don't have a particularly quick argument for well-definedness, but it is indeed well-defined (i.e., does not depend on the particular diagonalization that you use). The rough reason for this is that if we have two diagonalizations A = PDP^-1 = QEQ^-1 then PD^rP^-1 and QE^rQ^-1 are both matrices with eigenvalues d_{j,j}^r and e_{j,j}^r, respectively, and corresponding eigenvectors p_j and q_j, respectively. Since diagonalizable matrices are completely determined by their eigenvalues and eigenvectors, it must be the case that PD^rP^-1 = QE^rQ^-1. And yep, this definition coincides with the A^r=exp(r log A) definition. Cheers!
Thanks for your great teaching. May I ask a question? How Can we comute A^(2^1/2)?
You can use the definition of matrix powers at 12:36 in the video. First, diagonalize A as A = PDP^-1. Then raise each diagonal entry of D to the power of 2^(1/2), giving you the matrix D^(2^1/2). Then multiply to get A^(2^1/2) = P*D^(2^1/2)*P^-1.
@@NathanielMath Thanks. But I means in general How Can we do it. Not for digonalizable matrix
Hi, is there a quick argument wich shows, that the definition given at around 13:00 for the power A^r of a matrix is well defined (does not depend on the chosen transformation matrix, i.e. the order of the eigenvalues)? Does this definition match with the definition A^r=exp(r log A), for the matrix logarithm / exponential? Thanks in advance :)
Good question! I don't have a particularly quick argument for well-definedness, but it is indeed well-defined (i.e., does not depend on the particular diagonalization that you use). The rough reason for this is that if we have two diagonalizations A = PDP^-1 = QEQ^-1 then PD^rP^-1 and QE^rQ^-1 are both matrices with eigenvalues d_{j,j}^r and e_{j,j}^r, respectively, and corresponding eigenvectors p_j and q_j, respectively. Since diagonalizable matrices are completely determined by their eigenvalues and eigenvectors, it must be the case that PD^rP^-1 = QE^rQ^-1.
And yep, this definition coincides with the A^r=exp(r log A) definition. Cheers!