# Showing det(C) = det(A)det(B)

• Ylle
In summary, the conversation discusses a proof for the equation det(C) = det(A)det(B), where A and B are matrices and C is a j x k-zeromatrix. Different methods are suggested, such as using induction or the definition of a determinant, and the idea of using elementary matrices is also mentioned. Ultimately, the suggestion to use permutations is considered the most intuitive approach.

## Homework Statement

where 0 is the j x k-zeromatrix.

Show that det(C) = det(A)det(B)

??

## The Attempt at a Solution

Got no clue. So I could really use a clue or some help here :)

Regards

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Hint: Use induction on the dimension of $$B$$, and expand the determinant along the bottom row.

Hmmm, not totally sure what you mean.
Induction, isn't that the thing with:
1 + 2 + 3 + 4 + ... + n = (n)(n+1)/2 ?

Not sure how I use that on the dimension of B tbh :?

You can also use the definition of a determinant, i.e. the expression involving a summaton over permutations. Then it is trivial.

Are you familiar with elementary matrices?

Count Iblis said:
You can also use the definition of a determinant, i.e. the expression involving a summaton over permutations. Then it is trivial.
You mean:
det(A) = a1jA1j+...+ajjAjj
det(B) = b1kB1k+...+bkkBkk

And then when I multiply det(A) and det(B), I get the same as if I were to do the determinant of C, which will be det(A)*det(B)-0*0=det(A)*det(B) ?

Or that won't work because they don't have the same dimensions ?

C.E said:
Are you familiar with elementary matrices?

A matrix that does the same to an already existing matrix, as if you were to row-operate ?

Ylle said:
You mean:
det(A) = a1jA1j+...+ajjAjj
det(B) = b1kB1k+...+bkkBkk

And then when I multiply det(A) and det(B), I get the same as if I were to do the determinant of C, which will be det(A)*det(B)-0*0=det(A)*det(B) ?

Or that won't work because they don't have the same dimensions ?

I mean this: If A is an $n\times n$ matrix then, by definition, we have:

$$\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{i,\pi(i)}$$

where the summation is over all permutations $\pi$ of the numbers $1\ldots n$

Count Iblis said:
I mean this: If A is an $n\times n$ matrix then, by definition, we have:

$$\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{i,\pi(i)}$$

where the summation is over all permutations $\pi$ of the numbers $1\ldots n$

Ok, that way...

But how does that help me :?
I can make det(A) and det(B) into a linear transformation, but if I do the same on C, won't I just get det(C) = A*B ?

Consider some arbitrary permutation $\pi$. Then consider the contribution to the determinant:

$$\operatorname{sign}(\pi)C_{1,\pi(1)}C_{2,\pi(2)}\ldots C_{n,\pi(n)}$$

Where $C_{r,s}$ means the element in the rth row and sth column of C. And n =j+k is the dimension of C.

Now unless the permutation $\pi$ permutes the numbers tranging from 1 to j amongst themselves, you will get zero. So, this means that such permutations $\pi$ can be decomposed as a product of permutations

$$\pi=\rho\sigma$$

where $\rho$ permutes the numbers ranging from 1 to j and acts as the identity on the numbers ranging from j+1 to j+k and $\sigma$ is a permutation that permutes the numbers ranging from j+1 to j+k while acting as the identity on the numbers ranging from 1 to j.

Then, if you use that the sign of a product of permutations is the product of the signs of the permutations, you are done.

Hmmm, I'm really sorry. But I'm not sure I understand you here :?
It sounds a bit confusing in my ears - even though it's probably not :S

Ylle said:
Hmmm, I'm really sorry. But I'm not sure I understand you here :?
It sounds a bit confusing in my ears - even though it's probably not :S

If you are going to choose elements from the first j rows of C, then unless you choose them from submatrix A you'll get zero, right? So, the values you must choose from rows 1 to j are restricted in the range from 1 to j. Therefore the permutation [itex]\pi[/tex] will map the range 1 to j to itself.

Hmmm, not totally sure what you mean.
Induction, isn't that the thing with:
1 + 2 + 3 + 4 + ... + n = (n)(n+1)/2 ?

Not sure how I use that on the dimension of B tbh :?

Induction is a proof technique that takes advantage of an important property of the integers, namely, well-ordering. It's often compared to pushing down a stack of dominoes. What you do is show that your result holds for some base case (for your exercise, this would be the case where $$B$$ is a one-by-one matrix), and then show that if it holds for the case corresponding to a natural number $$n$$ (in your case, $$B$$ would be an $$n \times n$$ matrix), then it necessarily holds for $$n + 1$$. Carrying this out in your case isn't difficult, if you're familiar with this kind of argument.

However, I actually like Count Iblis's idea better here. Although they're easy to crank out, I try to avoid induction proofs where possible, because they typically don't confer a very intuitive grasp of the theorem upon the reader (or writer). The proof using permutations gives an immediate and intuitive idea of why this result should be true.

## 1. How do you show that det(C) is equal to the product of det(A) and det(B)?

To show that det(C) is equal to det(A)det(B), we can use the property that the determinant of a matrix is equal to the product of its eigenvalues. We can also expand the determinant of C using the Laplace expansion and show that it is equal to the product of the determinants of A and B.

## 2. Can you explain the concept of determinants and how they relate to matrix multiplication?

Determinants are a mathematical concept used to describe certain properties of a square matrix. They are a numerical value that represents the scaling factor of the matrix, and can tell us information about the matrix such as its invertibility and its effect on the volume of a shape. The product of the determinants of two matrices is equal to the determinant of their product, which is why det(C) = det(A)det(B) when multiplying matrices.

## 3. Is there a visual or intuitive way to understand why det(C) = det(A)det(B)?

Yes, there is a visual way to understand this property. When we multiply two matrices, we are essentially combining their transformations. The determinant of a matrix represents the scaling factor of its transformation, so when we multiply two matrices, the resulting determinant is the product of their scaling factors. This is why det(C) = det(A)det(B) when multiplying matrices.

## 4. Are there any exceptions to the rule that det(C) = det(A)det(B)?

Yes, there are some exceptions to this rule. One exception is when one or both of the matrices are not square matrices. In this case, the determinant is not defined. Another exception is when the matrices are not compatible for multiplication, such as when the number of columns in the first matrix does not match the number of rows in the second matrix. In this case, the product does not exist and therefore the determinant cannot be calculated.

## 5. How is the property det(C) = det(A)det(B) used in real-world applications?

The property det(C) = det(A)det(B) is commonly used in solving systems of linear equations, which have many applications in fields such as engineering, physics, and economics. It is also used in calculating the inverse of a matrix, which is important in solving many mathematical problems and in computer graphics. Additionally, the concept of determinants has applications in areas such as quantum mechanics and cryptography.