# I Determinant of A^t A

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1. Jun 14, 2016

### tommyxu3

I have a problem of proving an identity about determinants. For $A\in M_{m\times n}(\mathbb{R}),$ a matrix with $m$ rows and $n$ columns, prove the following identity.

$$|\det(A^tA)|=\sum_{1\le j_1\le ... \le j_n \le m} (det(A_{j_1...j_n}))^2$$
where $A_{j_1...j_n}$ is the matrix whose $(i,k)$-entry is $a_{j,k},$ and $A^t$ is the transpose of $A.$

The example of this is given here to make clear:

$$\det ( \begin{pmatrix} 1 & 3 &5\\ 2 & 4 &6\\ \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}) = (\det\begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix})^2+ (\det\begin{pmatrix} 1 & 2 \\ 5 & 6 \\ \end{pmatrix})^2+ (\det\begin{pmatrix} 3 & 4 \\ 5 & 6 \\ \end{pmatrix})^2 .$$

The equation is clearly holds fro square matrix, but for general type I cannot solve...I try to prove it in induction from $m\times n$ to $(m+1)\times n$ but failed. This may be related to the notion of area (the given example is the area of a triangle on a plane).
Thanks for any ideas in advance!!!

2. Jun 14, 2016

### Staff: Mentor

I have trouble to understand the notation $A_{j_1 ... j_n}$. Your example is a $m \times n = 3 \times 2$ matrix, so possible arrays are $(1,1),(1,2),(1,3),(2,2),(2,3),(3,3).$ Could you explain, e.g. $A_{13}$ and $A_{22}$?

3. Jun 14, 2016

### tommyxu3

That's my error, and in my example you may see apparently. The correct version should be: $\sum_{1\le j_1<...<j_n\le n}.$

4. Jun 14, 2016

### Staff: Mentor

Still doesn't explain $A_{13}$. Normally such a notation denotes a diagonal or cofactor. What does it mean for a matrix, that isn't square?

5. Jun 14, 2016

### tommyxu3

It means the matrix$$\begin{pmatrix} 1 & 2 \\ 5 & 6 \\ \end{pmatrix},$$
like the notation above.
Besides I found another error that it should be: "where $A_{j_1...j_n}$ is the matrix whose $(i,k)$-entry is $a_{j_ik},$ and $A^t$ is the transpose of $A.$"

6. Jun 14, 2016

### Staff: Mentor

Sorry. Maybe I'm stupid or stubborn, possibly both. But if
$$A_{13} = \begin{bmatrix} 1 && 2 \\ 3 && 4 \\ 5 && 6 \end{bmatrix}_{13} = \begin{bmatrix} 1 && 2 \\ 5 && 6 \end{bmatrix},$$
$A_{13}$ means the second row is deleted? But your sum allows only the array $(1,2)$ since you corrected the summation boundary from $m = 3$ to $n=2$.
Or do you write the columns first and rows second? You have, as I am used to, a $(3 \times 2)-$matrix $A \in \mathbb{M}_{3 \times 2}(\mathbb{R}).$ Note that $A$ is the second factor in $\det (A^t A)$.

I try to understand the task and if it is possible at all. Matrices that aren't square matrices don't occur that often so I'm not used to them. Especially I'm asking myself how the formula would look like, if $A$ is simply a vector.

7. Jun 14, 2016

### tommyxu3

Oops, error again... My correction focused on changing $"\le"$ to $"<".$ The boundary is from $1$ to $m$ like initially.Thanks for your check again!! Do you have any ideas?

8. Jun 14, 2016

### Staff: Mentor

If $A^t = (a_1, \dots ,a_m)$ then $A$ has $m$ rows and $n=1$ column.
Thus $| \det(A^t A) | = | \sum_{j=1}^{j=m} a_j^2 |$ and $1 \leq j_1 \leq m$ is the only array of indices $j_k$. Therefore $A_j = a_j$ is the only meaningful way to define $A_j$ and the formula is valid. So, I assume the $j-$ array denotes the numbers of rows that are not to be eliminated.
Give me a second to check the left hand side determinant of your example. The general formula for determinants is a bit complicated to handle.

9. Jun 14, 2016

### tommyxu3

That's right what I want to convey.

10. Jun 14, 2016

### Staff: Mentor

Ok. Your example isn't a counterexample. Both sides sum up to $24$.
Induction is probably an ugly task. We have $m \times 1$ so the induction step would be the equation for $m \times (n+1)$ given the formula for $m \times n$. I have to think about a less uncomfortable way. I'm sure there is one.
Edit: It could certainly be done by calculating the determinant along the additional last column.

11. Jun 14, 2016

### Staff: Mentor

An idea is to make $A$ a square matrix by filling the additional positions with zeros. Only on the main diagonal have to be ones.

12. Jun 14, 2016

### tommyxu3

I have tried, and let me see later!! Thanks~~
But I remembered the remainder may make some troubles?

13. Jun 14, 2016

### mathwonk

14. Jun 15, 2016

### tommyxu3

Yes, and I say that the statement is true for square matrix. You may misunderstand the statement that I didn't require $A$ is square.

15. Jun 15, 2016

### tommyxu3

My another idea is since it holds for square, then do inductions from $m\times n$ to $(m+1)\times n$?
By the way, your edit meant the induction from$m\times n$ to $m\times (n+1)$ works?

16. Jun 15, 2016

### Staff: Mentor

It should work, because the statement appears to be true. I have taken this inductive step, because the basis case $m \times 1$ is obviously true and allows to be proven for any $m$ so no other induction is necessary. And expanding a determinant along only one column (the $1$ in $m \times (n+1)$) and applying the inductive hypothesis $(n \times m)$ to the sub-determinants of this expansion might keep the mess bounded.

If we take $A$ to be a $(1 \times n)-$matrix, then $A^t A$ is a $(n \times n)-$matrix of rank $1$, i.e. zero determinate. On the RHS we then get an empty matrix or empty sum, which is be definition $0$. So even in this extreme case it is true. Have you checked other matrices of determinant zero, just to see what happens?

I have also thought about other ways. Replacing the determinants by its complete formulas (weighted sum over all diagonal products) looks even more messy. The LHS might be controllable, but I have no idea how to interpret the RHS with its wiped-out rows.

17. Jun 15, 2016

### tommyxu3

Really... other expansions may take lots of time...anyway...

18. Jun 15, 2016

### Staff: Mentor

If you are allowed to interpret determinants by volumes of cubes, this might shorten the proof. The reduced matrices $A_{j_1 ... j_n}$ could then be interpreted as orthogonal projections on subspaces.

19. Jun 15, 2016

### mathwonk

in relation to fresh's idea to make the matrix square, what about adding (to the given 3x2 example) a third row which is a unit vector orthogonal to the other two rows? then the new product is 3x3 and equals the old product plus a "1" in the lower right corner, and zeroes elsewhere in the new row and column, so getting the same determinant. this should generalize to adding new rows that are orthogonal to the old rows and to each other and have length one. does this work?

oops, this makes the matrix square, but does not obviously prove the result.

Last edited: Jun 15, 2016
20. Jun 15, 2016

### mathwonk

isn't this related to the pythagorean theorem for areas? i.e. isn't the sum of the squares of those 2x2 determinants equal to the square of the area of the parallelogram spanned by the 2 rows in 3 space? i.e. as fresh(?) suggested, the sum of the squares of the areas of the projections of the parallelogram onto the 3 planes, equals the square of the area of the original parallelogram. i seem to recall this, maybe provable from looking at the coordinates of the cross product?

21. Jun 16, 2016

### tommyxu3

It is!~~
And finally I proved it by decomposing it to matrix unit~ I think the induction also look troublesome...haha