# Thread: Orthogonal Matrix

1. ## Orthogonal Matrix

Can anyone help me out with these questions:

Given: Matrix(P)=
0-4-1
0 3 7
1 0 0

Such That: P-1AP=D is diagonal arranged so that the (1,1)-entry of D is different from the other diagonal entries.

Find an orthogonal matrix Q so that transpose(Q)AQ=D.
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2)

Given: matrix(A)=

-14/25 48/25 0
48/25 14/25 0
0 0 0

Find a diagonal matrix D and an orthogonal matrix Q so that the given matrix A = Q D transpose(Q).

Plz Help

2. Originally Posted by yakuut
Can anyone help me out with these questions:

Given: Matrix(P)=
0-4-1
0 3 7
1 0 0

Such That: P-1AP=D is diagonal arranged so that the (1,1)-entry of D is different from the other diagonal entries.

Find an orthogonal matrix Q so that transpose(Q)AQ=D.
The question says that "(1,1)-entry of D is different from the other diagonal entries", but I think that in order to answer the question you need to know more than that, namely that the other two diagonal entries of D are the same, so that D is of the form $\begin{bmatrix}\lambda&0&0\\ 0&\mu&0\\ 0&0&\mu\end{bmatrix}$. In the absence of any information at all about the matrix A, you will surely need to make that assumption about D.

If $P^{-1}AP = D$ then $A = PDP^{-1}$. The idea is to write P in the form $P=QR$, where Q is orthogonal and R is a positive definite matrix that commutes with D. Then $A = PDP^{-1} = QRDR^{-1}Q^{\textsc{t}} = QDQ^{\textsc{t}}$, from which $Q^{\textsc{t}}AQ = D$.

The factorisation P = QR is called the polar decomposition of P. The positive definite factor R is the square root of $P^{\textsc{t}}P$. So start by calculating $P^{\textsc{t}}P = \begin{bmatrix}1&0&0\\ 0&25&25\\ 0&25&50\end{bmatrix}$. This has square root $R = \begin{bmatrix}1&0&0\\ 0&0&-5\\ 0&-5&-5\end{bmatrix}$. (Fortunately, this commutes with D, as you can easily check.) Then $Q = PR^{-1} = \begin{bmatrix}0&-\tfrac35&\tfrac45\\ 0&-\tfrac45&-\tfrac35\\ 1&0&0\end{bmatrix}$.

Originally Posted by yakuut
2)

Given: matrix(A)=

-14/25 48/25 0
48/25 14/25 0
0 0 0

Find a diagonal matrix D and an orthogonal matrix Q so that the given matrix A = Q D transpose(Q).
This is much easier. The diagonal entries of D are the eigenvalues of A, and the columns of Q are the associated normalised eigenvectors.

3. Thanks Opalg for ur efforts..but the solution for the first question didn't work
thanks anyways...