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Source: 2023 Aug taca-9

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#1 h Apr 9, 2025, 9:55 AM • 1 Y

Y by Euler_Gauss

Let matrix
$A=\begin{bmatrix} 1&1&1&1&1\\1&-1&1&-1&1\\?&?&?&?&?\\?&?&?&?&?\\?&?&?&?&?\end{bmatrix}\in\mathbb R^{5\times 5}$ satisfy $\text{tr} (AA^T)=28.$ Determine the maximum value of $\det A.$

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#2 h Apr 9, 2025, 10:56 AM

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With a computer, I find $72$ .
Assuming that the last three rows of $A$ are real.

This post has been edited 1 time. Last edited by loup blanc, Apr 9, 2025, 6:01 PM

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#3 h Apr 9, 2025, 11:34 PM

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$\bullet$ Let $A_{i,j}$ be the matrix obtained from $A$ removing the $i^{th}$ row and the $j^{th}$ column.
If $i\geq 3$ , then $\dfrac{\partial \det(A)}{\partial a_{i,j}}=(-1)^{i+j}\det(A_{i,j})$ .
The imposed condition is $\sum_{i\geq 3,j\geq 1}{a_{i,j}}^2=18$ ; this set is a compact of $\mathbb{R}^{15}$ ,
what implies that the required maximum is reached and -easy to see- is $>0$ .
Then an extremum satisfies (Kuhn-Tucker) there is $k$ s.t. for every $i\geq 3,j\geq 1$ , $(-1)^{i+j}\det(A_{i,j})=ka_{i,j}$ .
Thus $\det(A)=k\sum_j{a_{3,j}}^2=k\sum_j{a_{4,j}}^2=k\sum_j{a_{5,j}}^2$ .
Then $\sum_j{a_{3,j}}^2=\sum_j{a_{4,j}}^2=\sum_j{a_{5,j}}^2=6$ and $\det(A)=6k$ .
$\bullet$ let $A=\begin{pmatrix}U_{2,5}\\V_{3,5}\end{pmatrix}$ , where $U$ is known and $V$ unknown. $AA^T=\begin{pmatrix}UU^T&UV^T\\VU^T&VV^T\end{pmatrix}$ is symmetric $>0$ and $tr(VV^T)=18$ (*).
$\det(A)^2=\det(UU^T)\det(VV^T-VU^T(UU^T)^{-1}UV^T)=24\det(VV^T-VU^T(UU^T)^{-1}UV^T)$ .
Note that $VU^T(UU^T)^{-1}UV^T\in M_3$ is symmetric $\geq 0$ .
If I'm not mistaken, $Z=VV^T-VU^T(UU^T)^{-1}UV^T$ is -as $AA^T$ - symmetric $> 0$ and $VV^T$ is invertible.
We seek the maximum of $\det(Z)$ . Under the condition (*), $\max(\det(VV^T))=6^3$ .
Finally, if we find $V$ s.t. $VV^T=6I_3$ and $VU^T=0$ , then we obtain $\max\det(A)^2=24\det(6 I_3)=24\times 6^3$
and $\max(\det(A))=72$ and $k=12$ .
$\textbf{Conclusion.}$ It remains to show that there is $V$ s.t. $VV^T=6I_3$ and $VU^T=0_{3,2}$ .

This post has been edited 2 times. Last edited by loup blanc, Apr 9, 2025, 11:36 PM

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#4 h Apr 10, 2025, 6:36 AM • 1 Y

Y by loup blanc

Nice problem! What is taca-9?

A somewhat different approach: The trace condition means that $g(A) = \sum_{i,j} a_{i,j}^2 = 28$ . By compactness the maximum is attained (as pointed out in the post above).

Consider the determinant as a function of the remaining row vectors $u,v,w$ of $A$ . Since the determinant is linear as a function of every row, $\det(A)$ can not be maximal if one of the rows is the $o$ -vector. Also $\det(A)$ can be maximal only if the Euclidean norm of $u,v,w$ are equal: If $|u| > |v| > 0$ , then choose $c > 1$ such that $\frac 1 c |u| = c |v|$ and replace $u$ by $\frac 1 c u$ and $v$ by $cv$ . This does not change the value of the determinant, but it is easy to see that it decreases the value of $g(A)$ . (If you write this out, this boils down to the AM-GM-inequality.) In conclusion the three remaining rows have norm $\sqrt 6$ (and this takes care of the $g(A) = 0$ condition).

Finally: For $u,v,w$ with fixed lengths the (absolute value of the) determinant is maximal iff $u,v,w$ are orthogonal to each other and to the first two row vectors. (This follows geometrically by interpreting the deterimant as the volume of the parallelepiped spanned by the row vectors. (It is clear that $u,v,w$ with this property exists and by swapping two rows we can make the determinant positive.) Calculating the area of the parallelogramm spanned by the first two row vectors gives $\sqrt{5^2-1^2} = \sqrt {24}$ . Thus the maximal value of the deterimant is $\sqrt{24} \sqrt 6^3 = 72$ (which matches the answer given above). Note that there are many matrices such that equality occurs.

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#5 h Apr 12, 2025, 10:59 AM

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The end of my post #3.
That follows is a matrix that satisfies the relations required in the conclusion.
$V=\begin{pmatrix}-\sqrt{2}&-1&\sqrt{2}&1&0\\-\sqrt{2}&1&0&-1&\sqrt{2}\\0&-1&-\sqrt{2}&1&\sqrt{2}\end{pmatrix}$ .

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