2
$\begingroup$

Consider the positive definite and symmetric matrix

$$A = \begin{pmatrix} 1 & 2 & 0 \\ 2 & 6 & -1 \\ 0 & -1 & 1 \end{pmatrix}$$

Find a decomposition with unipotent $U \in \mbox{Mat} (3,3,\mathbb{R})$ and a diagonal matrix $D \in \mbox{Mat} (3,3,\mathbb{R})$ such that $$U^tAU = D$$

I struggle with this task since this is different in the sense that D does not contain the eigenvalues, so it is not just an application of the spectral theorem for symmetric real matrices. Instead, it is $\mbox{diag}(\delta_1, \frac{\delta_2}{\delta_1}, \dots, \frac{\delta_n}{\delta_{n-1}})$ where $\delta_n$ stands for the $n$-th principal minor of the matrix $A$. Thus, to calculate $D$ is easy. However, how do I find those unipotent transformation matrices $U$?

$\endgroup$
1
$\begingroup$

Let us look for matrices $\mathrm U$ and $\mathrm D$ of the form

$$\mathrm U = \begin{bmatrix} 1 & u_1 & u_2\\ 0 & 1 & u_3\\ 0 & 0 & 1\end{bmatrix} \qquad \qquad \qquad \mathrm D = \begin{bmatrix} d_1 & 0 & 0\\ 0 & d_2 & 0\\ 0 & 0 & d_3\end{bmatrix}$$

Using SymPy:

>>> u1, u2, u3 = symbols('u1 u2 u3')
>>> d1, d2, d3 = symbols('d1 d2 d3')
>>> U = Matrix([[1, u1, u2], [0, 1, u3], [0, 0, 1]])
>>> U
[1  u1  u2]
[         ]
[0  1   u3]
[         ]
[0  0   1 ]
>>> D = diag(d1,d2,d3)
>>> D
[d1  0   0 ]
[          ]
[0   d2  0 ]
[          ]
[0   0   d3]
>>> A = Matrix([[1, 2, 0], [2, 6, -1], [0, -1, 1]])
>>> A
[1  2   0 ]
[         ]
[2  6   -1]
[         ]
[0  -1  1 ]
>>> D - (U.T * A * U)
[  d1 - 1                 -u1 - 2                                   -u2 - 2*u3                     ]
[                                                                                                  ]
[ -u1 - 2       d2 - u1*(u1 + 2) - 2*u1 - 6              -u2*(u1 + 2) - u3*(2*u1 + 6) + 1          ]
[                                                                                                  ]
[-u2 - 2*u3  -u1*(u2 + 2*u3) - 2*u2 - 6*u3 + 1  d3 - u2*(u2 + 2*u3) - u3*(2*u2 + 6*u3 - 1) + u3 - 1]

We have $6$ equations in $6$ unknowns. The solution is

$$(u_1, u_2, u_3) = \left(-2,-1,\frac 12\right) \qquad \qquad \qquad (d_1, d_2, d_3) = \left(1,2,\frac 12\right)$$

Using SymPy to verify:

>>> U = Matrix([[1, -2, -1], [0, 1, 0.5], [0, 0, 1]])
>>> U
[1  -2  -1 ]
[          ]
[0  1   0.5]
[          ]
[0  0    1 ]
>>> U.T * A * U
[1  0   0 ]
[         ]
[0  2   0 ]
[         ]
[0  0  0.5]
>>> (U - eye(3))**3
[0  0  0]
[       ]
[0  0  0]
[       ]
[0  0  0]
$\endgroup$
  • $\begingroup$ Thanks Rodrigo de Azevedo! But isn't there some kind of simplified method for calculation of the U matrix? Like, for example, the transformation matrix that contains the eigenvectors in the diagonalizatiation case? $\endgroup$ – Taufi Jul 20 '16 at 15:39
  • $\begingroup$ @Taufi I don't know. This is the first time I work with unipotent matrices. $\endgroup$ – Rodrigo de Azevedo Jul 20 '16 at 15:40
  • $\begingroup$ @Taufi The leading principal minors of $A$ are $1, 2, 1$. Hence, the diagonal of $D$ should be, according to your statement, $1,2,\frac 12$, which is what I obtained using SymPy. $\endgroup$ – Rodrigo de Azevedo Jul 20 '16 at 16:03
  • $\begingroup$ Yes, that is true. The diagonal matrix is calculated pretty fast. But is there some formula as for the diagonal matrix or just your approach with a system of linear equations? $\endgroup$ – Taufi Jul 20 '16 at 16:13
0
$\begingroup$

Note that since $A$ is positive definite, it has a Cholesky decomposition $A = LL^T$, where $L$ is an lower-triangular matrix. It follows that $L^{-1}$ is also lower-triangular.

Since $L^{-1}$ is lower triangular, there exists a diagonal matrix $M$ such that $ML^{-1} = U^T$ has $1$s on the diagonal. Since $U$ is an upper-triangular matrix with $1$s on the diagonal, it must be unipotent. Moreover, we have $$ U^TAU = (ML^{-1})(LL^T)(ML^{-1})^T = M(LL^{-1})(LL^{-1})^T M^T = MM^T = M^2 $$ Thus, we have an algorithm to calculate $U$, and the entries of the diagonal matrix $D$ are the squares of the diagonal entries of $L$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.