- Whats the intuition for ABA in linear algebra?
If we have a matrix B B and would like to re-express the transformation of multiplication by B B in another basis, standard linear algebra tells us that the matrix expression of the same transformation in the new basis is a similar matrix: Bnew = ABA−1 B new = A B A 1 Here the matrix A A is called a change of basis matrix
- linear algebra - How does one prove that $ABA^ {−1} = B$ given that A . . .
How does one prove that ABA−1 = B A B A 1 = B given that A is an invertible matrix? [closed] Ask Question Asked 7 years, 5 months ago Modified 6 years, 6 months ago
- Square matrices, ABA = A? - Mathematics Stack Exchange
Square matrices, ABA = A? Ask Question Asked 8 years, 1 month ago Modified 8 years, 1 month ago
- linear algebra - $X=ABA^T$, X, A are matrix and need to solve matrix B . . .
X = ABAT X = A B A T, X is a square matrix and A is rectangular matrix and I need to solve matrix B by enforcing it as a diagonal matrix For non-singular matrix we can write B as A+XA+T A + X A + T, but B gets to be non-diagonal matrix How can I get B as diagonal matrix (Eigen decomposition of X gives different values of A)
- linear algebra - Prove that $ABA^T$ is symmetric when $A$ and $B$ are . . .
I have been learning about matrix symmetry and came up with a question that I can't seem to prove The idea is that the product of ABAT A B A T is a symmetric matrix
- Is there a name for the matrix operation $ABA^t$
I know in group theory, the operation ABA−1 A B A 1, i e the element A multiplied by the element B, and then multiplied by the inverse of A, is called conjugation When dealing with matrices, there is a similar operation that happens frequently: ABAt A B A t, where instead of multiplying by the inverse of A, you multiply by the transpose of A Does this operation or the resulting matrix
- What is the term for $ABA^{-1}$ - Mathematics Stack Exchange
If A, B A, B are some matrices such that their products are defined, is there a term to describe pre-multiplication by A A and post-multiplication by its inverse (ABA−1 A B A 1)
- To find $B$ for every $A$: $ABA=A$ - Mathematics Stack Exchange
Show that for every $n \\times n$ matrix $A$ there exists a matrix $B$ such that $ABA=A$
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