# Properties of Determinants

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

This is daunting.

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Homework Helper
Gold Member
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

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

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Homework Helper
Gold Member
As for the solution:

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

Nusc
omg that's it. i hate this course

Homework Helper
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

Nusc
I just have to get used to the proving techniques

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

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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?

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Palindrom
Nusc said:
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

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

Palindrom
Nusc said:
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 them together to show that they are square?

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!

Nusc
Oops typo, thanks.

Nusc
Palindrom said:
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!

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

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Palindrom
Nusc said:
The r in A is the jth column and in B it's the ith row. So when you say m=n, are you referring 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.

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.

Nusc
Yeah thanks, but I will be back with more

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).

Nusc
Haha what?