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Diagonalization Symmetric Matrices and QR factorization

Theorem: If AA is a symmetric matrix, then eigenvectors associated to distinct eigenvalues are orthogonal.

Definition: A square matrix PP with orthonormal columns is called an orthogonal matrix.

Theorem: If PP is a square orthonormal matrix then PP is invertible and the inverse is the transpose - P1=PtP^{-1}=P^t.

Proof: Let P=[p1    pn]P=[p_1\; \ldots\; p_n]. Then the ijijth entry of PtPP^tP is pipjp_i\cdot p_j which is 1 if i=ji=j and is 00 if iji\neq j.

Questions:

  • What is the determinant of an orthogonal matrix? What does it tell you geometrically?

  • Is every matrix with determinant ±1\pm 1 orthogonal?

  • How would you invert an orthogonal matrix?

Orthogonally Diagonalizable Matrices

Definition: A square matrix AA is orthogonally diagonalizable if there exists an orthogonal matrix PP and a diagonal matrix DD such that A=PDP1=PDPtA=PDP^{-1}=PDP^t.

(give an example in class.)

Theorem: (Spectral Theorem) A matrix AA is orthogonally diagonalizable if and only if AA is symmetric.

It is easy to see that if AA is orthogonally diagonalizable then AA is symmetric. The converse is harder and won't be proved in this class.

(Work out example 3 on page 354)

QR Factorization

Theorem: (QR factorization) Let A=[a1    amA=[a_1\;\ldots\;a_m be an n×mn\times m matrix with linearly independent columns. Then AA can be factored as A=QRA=QR where QQ is a n×mn\times m matrix with orthonormal columns and RR is an m×mm\times m matrix with nonnegative diagonal.

See the book for a full proof.

The matrix Q=[q1    qm]Q=[q_1\; \ldots\; q_m] is obtained by the Gram-Schmidt process. We can then obtain RR by computing QtA=RQ^tA=R. This RR will be upper triangular but the entries on the diagonal could be negative. But we can fix it.

This is useful for solving linear systems. Pivot manipulations can cause significant roundoff errors.

If AA is invertible and Ax=bAx=b and A=QRA=QR, then QRx=bQRx = b so Rx=QtbRx=Q^tb. This is now a triangular system which can be solved with backsubstituion.

(Work out example 4 on page 356)