(New page: '''''Statement: I am going to derive through a series of statements that transposing a matrix does NOT change its determinant.''''' First we will start with a 2x2 matrix as follows: '''...)
 
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c & d \end{bmatrix}</math>  
 
c & d \end{bmatrix}</math>  
  
So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A^T (the =transpose).
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So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A^T (the transpose).
  
 
A^T=  
 
A^T=  
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Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A^T). Simple enough...
 
Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A^T). Simple enough...
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 +
Now, we will use the power of induction to make some powerful assumptions, which will be proven in a bit.
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 +
Lets assume this is true for all cases of nxn...
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So now assume we have a nxn matrix called B:
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 +
Then we can say that det(B)=det(B^T)
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 +
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'''''I now propose a question that is food for thought for the rest of this derivation.
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Question: Is it true for all (n+1)x(n+1) matrices as well??'''''
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We shall now see...
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Let a (n+1)x(n+1) matrix called A=
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<math>\begin{bmatrix}
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a11 & a12 & a13 & a1m \\
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a21 & a22 & a23 & a2m \\
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a31 & a32 & a33 & a3m \\
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am1 & am2 & am3 & amm \\
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 +
\end{bmatrix}</math>

Revision as of 11:00, 7 December 2010

Statement: I am going to derive through a series of statements that transposing a matrix does NOT change its determinant.

First we will start with a 2x2 matrix as follows:


Let the 2x2 matrix A=

$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $

So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A^T (the transpose).

A^T=

$ \begin{bmatrix} a & c \\ b & d \end{bmatrix} $

So, det(A^T)=ad-cb.

Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A^T). Simple enough...

Now, we will use the power of induction to make some powerful assumptions, which will be proven in a bit.

Lets assume this is true for all cases of nxn...

So now assume we have a nxn matrix called B:

Then we can say that det(B)=det(B^T)


I now propose a question that is food for thought for the rest of this derivation.

Question: Is it true for all (n+1)x(n+1) matrices as well??

We shall now see...

Let a (n+1)x(n+1) matrix called A=

$ \begin{bmatrix} a11 & a12 & a13 & a1m \\ a21 & a22 & a23 & a2m \\ a31 & a32 & a33 & a3m \\ am1 & am2 & am3 & amm \\ \end{bmatrix} $

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Ryne Rayburn