The Column Space of A Contains All Vectors Ax
The rightway of a Matrix multiply a Vector
\[A=\begin{bmatrix} 2 & 1 & 3\\ 3 & 1 & 4\\ 5 & 7 & 12\\ \end{bmatrix} , \boldsymbol x = \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix}\] \[A \boldsymbol x = \begin{bmatrix} 2 & 1 & 3\\ 3 & 1 & 4\\ 5 & 7 & 12\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} = x_1 \begin{bmatrix} 2\\ 3\\ 5\\ \end{bmatrix} + x_2 \begin{bmatrix} 1\\ 1\\ 7\\ \end{bmatrix} + x_3 \begin{bmatrix} 3\\ 4\\ 12\\ \end{bmatrix}\]-
The combination of vector produce a vector.
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Thinking matrix as a whole thing.
Think about all combinations of the columns of $A$
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All $A \boldsymbol x$ give us a big banch of vectorx, that collection of vectors is called the column space of $A$. It’s a space, in other words, that’s the keyword there, the column space of $A$.
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In this case, we got a plane.
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The below column space is a line ($A$ is rank 1 matrix):
- The rank is sort of the dimension of the column space
Matrices with two factor
- Basis for the column space ($C$)
- The column rank is 2
- column rank = row rank = 2, and why?
What’s the row rank, what’s the row space?
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All combinations of the row is row space
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Two ways to get the row space
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transpose the matrix, and to get row space, we need to get transpose matrix’s column space
$R(A) = C(A^T)$
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the basis for the row space is $R= \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1\\ \end{bmatrix}$
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the conditions to be basis
- Independent
- the combinations produce all the rows
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Range?
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The factorization of matrix
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$A = CR$
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If you have a giant matrix, like size 10 to the 5th, you can’t put that into fast memory. It’s a mess. How do you deal with a matrix of size 10 to the 5th, when you cannot deal with all the entries?
How do you sample a matrix?
you want to get some typical columns.
$A\boldsymbol x, x = rand(m, 1)$
$A(BC\boldsymbol x)$
$A = CUR^\prime$, $R^\prime$ is the rows of $A$
Matrix multiply a Matrix
- Dot product
- New way
