How to find examples of matrices $A$ such that $AA^T$ equals a given matrix?

Question

I want to find examples of square matrices $A$ (and if possible, a general form) which satisfy the following property:

$$AA^{T} = \frac{1}{4} \left[\begin{matrix} 15 & 9 & 5 & -3 \\ 9 & 15 & 3 & -5 \\ 5 & 3 & 15 & -9 \\ -3 & -5 & -9 & 15 \end{matrix}\right]$$

What would a systematic way to go about this?

P.S: The matrix on the right hand side is Hermitian.

Answer 1

Since $B=AA^T$ is symmetric, by eigenvalues and eigenvectors, we can find $Q$ orthogonal and $\Lambda$ diagonal such that

$$B=Q\Lambda Q^T$$

and if $B$ is positive definite we have

$$B=Q\Lambda Q^T=(Q\Lambda^{1/2})(Q\Lambda^{1/2})^T=AA^T$$

Answer 2

You would want this version:

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 3 }{ 5 } & 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & 0 & 1 & 0 \\ - \frac{ 1 }{ 5 } & - \frac{ 1 }{ 3 } & - \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 15 & 0 & 0 & 0 \\ 0 & \frac{ 48 }{ 5 } & 0 & 0 \\ 0 & 0 & \frac{ 40 }{ 3 } & 0 \\ 0 & 0 & 0 & \frac{ 128 }{ 15 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 5 } & \frac{ 1 }{ 3 } & - \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & - \frac{ 3 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 15 & 9 & 5 & - 3 \\ 9 & 15 & 3 & - 5 \\ 5 & 3 & 15 & - 9 \\ - 3 & - 5 & - 9 & 15 \\ \end{array} \right) $$

Since $D$ is positive definite, create a matrix $F$ with entries $\sqrt d$ to get $Q^T F^T F Q = (FQ)^T (FQ) = H$ after deleting the $D.$ Let's see, you had a factor of $1/4,$ so $ (FQ/2)^T (FQ/2) = H/4$

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrrr} 15 & 9 & 5 & - 3 \\ 9 & 15 & 3 & - 5 \\ 5 & 3 & 15 & - 9 \\ - 3 & - 5 & - 9 & 15 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrrr} 15 & 9 & 5 & - 3 \\ 9 & 15 & 3 & - 5 \\ 5 & 3 & 15 & - 9 \\ - 3 & - 5 & - 9 & 15 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 3 }{ 5 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 3 }{ 5 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 5 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrr} 15 & 0 & 5 & - 3 \\ 0 & \frac{ 48 }{ 5 } & 0 & - \frac{ 16 }{ 5 } \\ 5 & 0 & 15 & - 9 \\ - 3 & - \frac{ 16 }{ 5 } & - 9 & 15 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrrr} 1 & 0 & - \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrr} 1 & - \frac{ 3 }{ 5 } & - \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 5 } & \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrr} 15 & 0 & 0 & - 3 \\ 0 & \frac{ 48 }{ 5 } & 0 & - \frac{ 16 }{ 5 } \\ 0 & 0 & \frac{ 40 }{ 3 } & - 8 \\ - 3 & - \frac{ 16 }{ 5 } & - 8 & 15 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrrr} 1 & 0 & 0 & \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrr} 1 & - \frac{ 3 }{ 5 } & - \frac{ 1 }{ 3 } & \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 5 } & \frac{ 1 }{ 3 } & - \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrr} 15 & 0 & 0 & 0 \\ 0 & \frac{ 48 }{ 5 } & 0 & - \frac{ 16 }{ 5 } \\ 0 & 0 & \frac{ 40 }{ 3 } & - 8 \\ 0 & - \frac{ 16 }{ 5 } & - 8 & \frac{ 72 }{ 5 } \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrrr} 1 & - \frac{ 3 }{ 5 } & - \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 5 } & \frac{ 1 }{ 3 } & - \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrrr} 15 & 0 & 0 & 0 \\ 0 & \frac{ 48 }{ 5 } & 0 & 0 \\ 0 & 0 & \frac{ 40 }{ 3 } & - 8 \\ 0 & 0 & - 8 & \frac{ 40 }{ 3 } \\ \end{array} \right) $$

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$$ E_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{ 3 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrr} 1 & - \frac{ 3 }{ 5 } & - \frac{ 1 }{ 3 } & - \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & \frac{ 3 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 5 } & \frac{ 1 }{ 3 } & - \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & - \frac{ 3 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrr} 15 & 0 & 0 & 0 \\ 0 & \frac{ 48 }{ 5 } & 0 & 0 \\ 0 & 0 & \frac{ 40 }{ 3 } & 0 \\ 0 & 0 & 0 & \frac{ 128 }{ 15 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 3 }{ 5 } & 1 & 0 & 0 \\ - \frac{ 1 }{ 3 } & 0 & 1 & 0 \\ - \frac{ 1 }{ 5 } & \frac{ 1 }{ 3 } & \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 15 & 9 & 5 & - 3 \\ 9 & 15 & 3 & - 5 \\ 5 & 3 & 15 & - 9 \\ - 3 & - 5 & - 9 & 15 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 3 }{ 5 } & - \frac{ 1 }{ 3 } & - \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & \frac{ 3 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 15 & 0 & 0 & 0 \\ 0 & \frac{ 48 }{ 5 } & 0 & 0 \\ 0 & 0 & \frac{ 40 }{ 3 } & 0 \\ 0 & 0 & 0 & \frac{ 128 }{ 15 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 3 }{ 5 } & 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & 0 & 1 & 0 \\ - \frac{ 1 }{ 5 } & - \frac{ 1 }{ 3 } & - \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 15 & 0 & 0 & 0 \\ 0 & \frac{ 48 }{ 5 } & 0 & 0 \\ 0 & 0 & \frac{ 40 }{ 3 } & 0 \\ 0 & 0 & 0 & \frac{ 128 }{ 15 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 5 } & \frac{ 1 }{ 3 } & - \frac{ 1 }{ 5 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 1 & - \frac{ 3 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 15 & 9 & 5 & - 3 \\ 9 & 15 & 3 & - 5 \\ 5 & 3 & 15 & - 9 \\ - 3 & - 5 & - 9 & 15 \\ \end{array} \right) $$

Answer 3

Looking again, this seems to be a contest type question. With a little trickery, one may find the eigenvalues entirely by hand, without attempting any 4 by 4 determinant. MORE TO COME

First, multiply by $4,$ the fraction can be dealt with later. Next, multiply one left and right by the orthogonal matrix (its own transpose) $$ \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right) $$

The result is a matrix in 2 by 2 blocks, $$ M = \left( \begin{array}{cc} 3A & A \\ A & 3A \end{array} \right) $$ where $$ A = \left( \begin{array}{cc} 5 & 3 \\ 3 & 5 \end{array} \right) $$ The eigenvalues of this are $2,8.$ We can construct eigenvectors for the 4 by 4 $M$ above with no trouble. If $v$ has eigenvalue $2,$ then as eigenvectors for my $M$ above, $$ \left( \begin{array}{c} v \\ v \end{array} \right) $$ has eigenvalue $8$ while $$ \left( \begin{array}{c} v \\ -v \end{array} \right) $$ has eigenvalue $4.$ If $w$ has eigenvalue $8,$ then as eigenvectors for my $M$ above, $$ \left( \begin{array}{c} w \\ w \end{array} \right) $$ has eigenvalue $32$ while $$ \left( \begin{array}{c} w \\ -w \end{array} \right) $$ has eigenvalue $16.$

So, my $M$ has eigenvalues $4,8,16,32.$ One may use the $M$ eigenvectors to reconstruct eigenvectors for the original matrix, or start over. Including the $1/4$ fraction, the matrix in the question has eigenvalues $1,2,4,8.$