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Prove that, if [tex]AA^T = A^TA = I_n[/tex], then [tex]\det{A} = \pm 1[/tex].
This is daunting.
This is daunting.
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Just put the 'T' out of the brackets at the end (that is, over all of: cA).Nusc said:Prove that any scalar multiple of a symmetric matrix is symmetric.
Let [tex]A = (a_i_j)[/tex]
Since [tex]A[/tex] is symmetric, [tex]A = A^T[/tex].
Then [tex]A^T = (a_i_j)[/tex].
Therefore, [tex](cA) = c(A) = c(A^T) = (cA^T)[/tex]
Was it necessary to show that [tex]A = (a_i_j)[/tex] and [tex]A^T = (a_i_j)[/tex]? Is [tex]A^T = (a_i_j)[/tex] even right? I can't express myself mathematically
If A+B is defined, then their sizes are similar- i.e., they're both mxn.Nusc said:Prove that, if A and B are two matrices such that A + B and AB are defined, then both A and B are square matrices.
- Let A be an m x r matrix and B an r x n matrix such that,
[tex]A_m_x_rB_r_x_n = (AB)_m_x_n[/tex]
- We know that the sum A + B of the two matrices is the m x n matrix
How do I express them together to show that they are square?
The r in A is the jth column and in B it's the ith row. So when you say m=n, are you refering to the r's?Palindrom said:Since AB is defined, as you yourself wrote, we must have m=n. (Because the product is an mxn*mxn, which is only defined when m=n).
Both matrices are therefor nxn, square matrices!
Yes, I was referring to the 'r's.Nusc said:The r in A is the jth column and in B it's the ith row. So when you say m=n, are you refering to the r's?
And if I were to prove that any scalar multiple of a diagonal matrix is a diagonal matrix, how is that different from, say, letting [tex] A = (a_i_j) [/tex] be any m x n matrix and c any real number?
A diagonal matrix is a square matrix that all of its nonzero entries are on the diagonal.
Then [tex] cA = c(a_i_j) = (ca_i_j)[/tex] but it may not be diagonal.