Linear Algebra question, what is the index of a column in a matrix?

[crowz]

As the title says, I'm trying to help a friend by explaining some problems, But a certain problem but i am just stuck on a point where it mentions "i" is the index of a particular column in a 4x4 matrix.
I cannot move on untill I know what they are refering to by index of a column .
Exact Line: "Let i be the index of the first column of M without a pivot"

Can some one point out what the index of a column is? for example [ 0 1 4 7] if that is a matrix,
..................................................................................................[ 3 5 8 1]
say index of column 3, is it (3,1) and (3,2) or is it 4 and 8 or what exactly?

 

[DrPhil]

As the title says, I'm trying to help a friend by explaining some problems, But a certain problem but i am just stuck on a point where it mentions "i" is the index of a particular column in a 4x4 matrix.
I cannot move on untill I know what they are refering to by index of a column .
Exact Line: "Let i be the index of the first column of M without a pivot"

Can some one point out what the index of a column is? for example [ 0 1 4 7] if that is a matrix,
..................................................................................................[ 3 5 8 1]
say index of column 3, is it (3,1) and (3,2) or is it 4 and 8 or what exactly?

The index of a column is just the column number. That is, the index of the 3rd column is 3.

Well, as an old Fortran programmer I can say that! If you program in C (or C++), you will usually have array indices starting with 0 instead of 1. So in program jargon, the index of the 3rd column might be 2. The index is a pointer .. it tells you where to find the column of interest.

 

[crowz]

Thank you for the reply
that would make the problem a bit odd, I might as well link the question:


My feelings is this is supposed to support that v is not equal to zero, but Mv = 0
v was this for me, im putting it as a row now but its a column [-1, 3, -1,0]
Mv was [0, 3, 10, 0] for me

if possible could you assist me with this one?

 

[crowz]

Any idea Dr.Phil?

 

[DrPhil]

Thank you for the reply
that would make the problem a bit odd, I might as well link the question:
View attachment 3268

My feelings is this is supposed to support that v is not equal to zero, but Mv = 0
v was this for me, im putting it as a row now but its a column [-1, 3, -1,0]
Mv was [0, 3, 10, 0] for me

if possible could you assist me with this one?

Lets see if I can add anything. I will use "^T" to indicate transposing rows to columns, and "_" for subscript.

i=3 is the first column without a pivot, and C_3 = [-1 3 0 0]^T
Your 3rd basis vector is b_3 = [0 0 1 0]^T
v = C_3 - b_3 = [-1 3 -1 0]^T
and M×v = [0 0 0 0]^T

Hmmm - doesn't look like what you got! Did I misinterpret multiplying the matrix times a column vector?
Or did I stumble onto the conclusion you wanted to reach?

 

[crowz]

Thank you, yes that is the correct answer, I must have made a mistake in multiplying M and v, can u go explain the proccess there?

 

[daon2]

In general, \(\displaystyle Mb_j = C_j\) for any column index \(\displaystyle j\) and \(\displaystyle j^{th}\) standard basis vector, so

\(\displaystyle M(C_i-b_i) = MC_i - Mb_i =MC_i-C_i = (M-I)C_i\)

 

[DrPhil]

Thank you, yes that is the correct answer, I must have made a mistake in multiplying M and v, can u go explain the proccess there?

Ordinary matrix multiplication: take the dot product of each row of M times v to get the corresponding element of the result.

\(\displaystyle \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 \end{bmatrix} \times \begin{bmatrix} -1 \\ 3 \\ -1 \\ 0 \end{bmatrix}
= \begin{bmatrix} (1)(-1)+(0)(3)+(-1)(-1)+(0)(0) \\ \cdot\cdot\cdot \\ \cdot\cdot\cdot \\ \cdot\cdot\cdot \end{bmatrix}
= \begin{bmatrix} 0 \\ \cdot \\ \cdot \\ \cdot \end{bmatrix} \)

Did you use columns instead of rows of M?