Linear Algebra, orthogonal columns and length
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Suppose A is a 3x3 matrix whose columns are orthogonal and the length (two-norm) of each column equals 4. Then what is $A^T*A$?
How would I even start proving this? I have to remain general such that my reasoning applies to any matrix of that form. I just cannot pick a random 3x3 orthogonal column matrix to see the outcome.
linear-algebra orthogonality orthogonal-matrices
edited Jan 18 at 13:51
Moo
5,65031020
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
$endgroup$
add a comment |
$begingroup$
Suppose A is a 3x3 matrix whose columns are orthogonal and the length (two-norm) of each column equals 4. Then what is $A^T*A$?
How would I even start proving this? I have to remain general such that my reasoning applies to any matrix of that form. I just cannot pick a random 3x3 orthogonal column matrix to see the outcome.
linear-algebra orthogonality orthogonal-matrices
edited Jan 18 at 13:51
Moo
5,65031020
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
$endgroup$
$begingroup$
Try to find out what is the entry $;1-1;$ in the product...Can you see that it is the square of the norm of the first column of $;A;$ ?
$endgroup$
– DonAntonio
Jan 18 at 13:55
add a comment |
$begingroup$
Suppose A is a 3x3 matrix whose columns are orthogonal and the length (two-norm) of each column equals 4. Then what is $A^T*A$?
How would I even start proving this? I have to remain general such that my reasoning applies to any matrix of that form. I just cannot pick a random 3x3 orthogonal column matrix to see the outcome.
linear-algebra orthogonality orthogonal-matrices
edited Jan 18 at 13:51
Moo
5,65031020
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
$endgroup$
Suppose A is a 3x3 matrix whose columns are orthogonal and the length (two-norm) of each column equals 4. Then what is $A^T*A$?
How would I even start proving this? I have to remain general such that my reasoning applies to any matrix of that form. I just cannot pick a random 3x3 orthogonal column matrix to see the outcome.
linear-algebra orthogonality orthogonal-matrices
linear-algebra orthogonality orthogonal-matrices
edited Jan 18 at 13:51
Moo
5,65031020
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
edited Jan 18 at 13:51
Moo
5,65031020
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
edited Jan 18 at 13:51
Moo
5,65031020
edited Jan 18 at 13:51
Moo
5,65031020
edited Jan 18 at 13:51
Moo
5,65031020
5,65031020
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
asked Jan 18 at 13:49
Wouter VandenputteWouter Vandenputte
1295
1295
$begingroup$
Try to find out what is the entry $;1-1;$ in the product...Can you see that it is the square of the norm of the first column of $;A;$ ?
$endgroup$
– DonAntonio
Jan 18 at 13:55
add a comment |
$begingroup$
Try to find out what is the entry $;1-1;$ in the product...Can you see that it is the square of the norm of the first column of $;A;$ ?
$endgroup$
– DonAntonio
Jan 18 at 13:55
$begingroup$
Try to find out what is the entry $;1-1;$ in the product...Can you see that it is the square of the norm of the first column of $;A;$ ?
$endgroup$
– DonAntonio
Jan 18 at 13:55
$begingroup$
Try to find out what is the entry $;1-1;$ in the product...Can you see that it is the square of the norm of the first column of $;A;$ ?
$endgroup$
– DonAntonio
Jan 18 at 13:55
add a comment |
1 Answer
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How you define the element $alpha_{ij}$ in the matrix $A^TA$? It's the scalar product of the $i$-th row of $A^T$ and the $j$-th column of $A$. But the $i$-th row of $A^T$ is the $i$-th column of $A$ (by definition of transpose). Thus $alpha_{ij}=0$ if $i neq j$ since the columns of $A$ are orthogonal, and $alpha_{jj}=16$. Therefore $A^TA=16I$ where $I$ is the identity matrix of order $3$
answered Jan 18 at 13:57
user289143user289143
1,069313
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1 Answer
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$begingroup$
How you define the element $alpha_{ij}$ in the matrix $A^TA$? It's the scalar product of the $i$-th row of $A^T$ and the $j$-th column of $A$. But the $i$-th row of $A^T$ is the $i$-th column of $A$ (by definition of transpose). Thus $alpha_{ij}=0$ if $i neq j$ since the columns of $A$ are orthogonal, and $alpha_{jj}=16$. Therefore $A^TA=16I$ where $I$ is the identity matrix of order $3$
answered Jan 18 at 13:57
user289143user289143
1,069313
$endgroup$
add a comment |
$begingroup$
How you define the element $alpha_{ij}$ in the matrix $A^TA$? It's the scalar product of the $i$-th row of $A^T$ and the $j$-th column of $A$. But the $i$-th row of $A^T$ is the $i$-th column of $A$ (by definition of transpose). Thus $alpha_{ij}=0$ if $i neq j$ since the columns of $A$ are orthogonal, and $alpha_{jj}=16$. Therefore $A^TA=16I$ where $I$ is the identity matrix of order $3$
answered Jan 18 at 13:57
user289143user289143
1,069313
$endgroup$
add a comment |
$begingroup$
How you define the element $alpha_{ij}$ in the matrix $A^TA$? It's the scalar product of the $i$-th row of $A^T$ and the $j$-th column of $A$. But the $i$-th row of $A^T$ is the $i$-th column of $A$ (by definition of transpose). Thus $alpha_{ij}=0$ if $i neq j$ since the columns of $A$ are orthogonal, and $alpha_{jj}=16$. Therefore $A^TA=16I$ where $I$ is the identity matrix of order $3$
answered Jan 18 at 13:57
user289143user289143
1,069313
$endgroup$
How you define the element $alpha_{ij}$ in the matrix $A^TA$? It's the scalar product of the $i$-th row of $A^T$ and the $j$-th column of $A$. But the $i$-th row of $A^T$ is the $i$-th column of $A$ (by definition of transpose). Thus $alpha_{ij}=0$ if $i neq j$ since the columns of $A$ are orthogonal, and $alpha_{jj}=16$. Therefore $A^TA=16I$ where $I$ is the identity matrix of order $3$
answered Jan 18 at 13:57
user289143user289143
1,069313
answered Jan 18 at 13:57
user289143user289143
1,069313
answered Jan 18 at 13:57
user289143user289143
1,069313
answered Jan 18 at 13:57
user289143user289143
1,069313
1,069313
add a comment |
add a comment |
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$begingroup$
Try to find out what is the entry $;1-1;$ in the product...Can you see that it is the square of the norm of the first column of $;A;$ ?
$endgroup$
– DonAntonio
Jan 18 at 13:55