- #1

Nusc

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Prove that, if [tex]AA^T = A^TA = I_n[/tex], then [tex]\det{A} = \pm 1[/tex].

This is daunting.

This is daunting.

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- Thread starter Nusc
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- #1

Nusc

- 760

- 2

Prove that, if [tex]AA^T = A^TA = I_n[/tex], then [tex]\det{A} = \pm 1[/tex].

This is daunting.

This is daunting.

Last edited:

- #2

quasar987

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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.

[tex]AA^T = A^TA = I_n[/tex]

then

[tex]\det{A} = \pm 1[/tex]

[tex]AA^T = A^TA = I_n[/tex]

then

[tex]\det{A} = \pm 1[/tex]

Last edited:

- #3

quasar987

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[tex]\det{AA^T} = \det{A}\det{A^T} = \det{A}^2 = \det{I_n} = 1 \Leftrightarrow \det{A} = \pm 1[/tex]

- #4

Nusc

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omg that's it. i hate this course

- #5

matt grime

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- #6

Nusc

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I just have to get used to the proving techniques

- #7

Nusc

- 760

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

Let [tex]A = (a_i_j)[/tex]

Since [tex]A[/tex] is symmetric, [tex]A = A^T[/tex].

Then [tex]A^T = (a_i_j)[/tex].

Therefore, [tex](cA) = c(A) = c(A^T) = (cA^T)[/tex]

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

Let [tex]A = (a_i_j)[/tex]

Since [tex]A[/tex] is symmetric, [tex]A = A^T[/tex].

Then [tex]A^T = (a_i_j)[/tex].

Therefore, [tex](cA) = c(A) = c(A^T) = (cA^T)[/tex]

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

Last edited:

- #8

Nusc

- 760

- 2

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,

[tex]A_m_x_rB_r_x_n = (AB)_m_x_n[/tex]

- 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?

- Let A be an m x r matrix and B an r x n matrix such that,

[tex]A_m_x_rB_r_x_n = (AB)_m_x_n[/tex]

- 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:

- #9

Palindrom

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Nusc said:Prove that any scalar multiple of a symmetric matrix is symmetric.

Let [tex]A = (a_i_j)[/tex]

Since [tex]A[/tex] is symmetric, [tex]A = A^T[/tex].

Then [tex]A^T = (a_i_j)[/tex].

Therefore, [tex](cA) = c(A) = c(A^T) = (cA^T)[/tex]

Was it necessary to show that [tex]A = (a_i_j)[/tex] and [tex]A^T = (a_i_j)[/tex]? Is [tex]A^T = (a_i_j)[/tex] 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).

- #10

Palindrom

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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,

[tex]A_m_x_rB_r_x_n = (AB)_m_x_n[/tex]

- 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!

- #11

Nusc

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Oops typo, thanks.

- #12

Nusc

- 760

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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 [tex] A = (a_i_j) [/tex] 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 [tex] cA = c(a_i_j) = (ca_i_j)[/tex] but it may not be diagonal.

I guess we assume that [tex] A = (a_i_j) [/tex] is a diagonal matrix?

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- #13

Palindrom

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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 [tex] A = (a_i_j) [/tex] 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 [tex] cA = c(a_i_j) = (ca_i_j)[/tex] 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.

Did I answer your question? :uhh:

- #14

Nusc

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Yeah thanks, but I will be back with more

- #15

Palindrom

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

- #16

Nusc

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Haha what?

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