# Properties of Determinants

1. Jul 11, 2005

### Nusc

Prove that, if $$AA^T = A^TA = I_n$$, then $$\det{A} = \pm 1$$.

This is daunting.

Last edited: Jul 11, 2005
2. Jul 11, 2005

### quasar987

try w/o the spaces btw the brackets and the "tex". you don't need to rewrite [ tex ] on both sides of the "=" u know.

$$AA^T = A^TA = I_n$$

then

$$\det{A} = \pm 1$$

Last edited: Jul 11, 2005
3. Jul 11, 2005

### quasar987

As for the solution:

$$\det{AA^T} = \det{A}\det{A^T} = \det{A}^2 = \det{I_n} = 1 \Leftrightarrow \det{A} = \pm 1$$

4. Jul 12, 2005

### Nusc

omg that's it. i hate this course

5. Jul 12, 2005

### matt grime

don't hate the course because it is daunting; see it as a challenge. as you see the answer is straight forward (i presume that's what the 'omg that's it' means), and all teh answers are like that. just be calm and check your notes for things

6. Jul 12, 2005

### Nusc

I just have to get used to the proving techniques

7. Jul 14, 2005

### Nusc

Prove that any scalar multiple of a symmetric matrix is symmetric.

Let $$A = (a_i_j)$$

Since $$A$$ is symmetric, $$A = A^T$$.

Then $$A^T = (a_i_j)$$.

Therefore, $$(cA) = c(A) = c(A^T) = (cA^T)$$

Was it necessary to show that $$A = (a_i_j)$$ and $$A^T = (a_i_j)$$? Is $$A^T = (a_i_j)$$ even right? I can't express myself mathematically

Last edited: Jul 14, 2005
8. Jul 14, 2005

### Nusc

Prove that, if A and B are two matrices such that A + B and AB are defined, then both A and B are square matrices.

- Let A be an m x r matrix and B an r x n matrix such that,
$$A_m_x_rB_r_x_n = (AB)_m_x_n$$

- We know that the sum A + B of the two matrices is the m x n matrix

How do I express this mathematically to show that they are square?

Last edited: Jul 14, 2005
9. Jul 14, 2005

### Palindrom

Just put the 'T' out of the brackets at the end (that is, over all of: cA).

10. Jul 14, 2005

### Palindrom

If A+B is defined, then their sizes are similar- i.e., they're both mxn.
Since AB is defined, as you yourself wrote, we must have m=n. (Because the product is an mxn*mxn, which is only defined when m=n).
Both matrices are therefor nxn, square matrices!

11. Jul 14, 2005

### Nusc

Oops typo, thanks.

12. Jul 14, 2005

### Nusc

The r in A is the jth column and in B it's the ith row. So when you say m=n, are you refering to the r's?

And if I were to prove that any scalar multiple of a diagonal matrix is a diagonal matrix, how is that different from, say, letting $$A = (a_i_j)$$ be any m x n matrix and c any real number?

A diagonal matrix is a square matrix that all of its nonzero entries are on the diagonal.

Then $$cA = c(a_i_j) = (ca_i_j)$$ but it may not be diagonal.

I guess we assume that $$A = (a_i_j)$$ is a diagonal matrix?

Last edited: Jul 14, 2005
13. Jul 14, 2005

### Palindrom

Yes, I was referring to the 'r's.
And I'm sure I got your question. It's true that for any matrix A=(aij) that keeps aij=0 for all i!=j, cA keeps the same thing for any scalar c. In the private case of square matrix, it shows that any scalar multiple of a diagonal matrix is, indeed, a diagonal matrix.

14. Jul 14, 2005

### Nusc

Yeah thanks, but I will be back with more

15. Jul 14, 2005

### Palindrom

Is that a threat?

Anyway, I'm off to bed. Be back in about 12 hours... (It's the middle of the night here).

16. Jul 14, 2005

Haha what?