Properties of Determinants

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  • #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.
 
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  • #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]
 
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  • #3
quasar987
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As for the solution:

[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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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
Nusc
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I just have to get used to the proving techniques
 
  • #7
Nusc
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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
 
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  • #8
Nusc
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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?
 
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  • #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
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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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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
Nusc
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Haha what?
 

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