Linear Algebra Part 1
May 3, 2016
This is the summary of the topics covered by Gilbert Strang in his lectures
Table of Contents:
Lectures 1-5
Covers the basic matrix concepts
- Matrix Row picture and Column picture
- Can we solve for \(Ax=b\) for all \(b\) ?. No, only for \(b\)’s in the column space of \(A\).
Four ways of matrix multiplication
- Dot product
- Column picture
- Row picture
- Column times Row
Gaussian Elimination
- Operates on the rows of matrix \(A\), finally puts the matrix in Row Reduced Echelon(REF) form
- Gaussian elimination leads to \(A = LU\) form, and \(A = LDU\) form
- Helps to put the elements of \(A\), in the Row Reduced Echelon Form (RREF) form
-
Can be used to solve for \(x\) in \(Ax = b\), by finding **_RREF([\(A\) |
\(b\)])** which results in **[\(I\) |
\(x\)]_** |
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Can be used to find the inverse of the matrix \(A\) by finding **_RREF[\(A\) |
\(I\)]** which results in **[\(I\) |
\(A^{-1}\)]_** |
Permutation matrix
- Number of permutation matrix \(A_{N*N} = N!\)
- \(P^T = P\) (where \(P^T\) is transpose(\(P\)))
- \(P P^T = I\) (Can you prove? Hint: use definition of matrix multiplication) * If the number of equations are less than number of variables, why we cannot solve the equation? * This is because rank of the matrix will be less than the number of variables we need to solve, so it will have infinite number of solutions * Use RREF(\(A\)) in matlab/octave and check
- \(A A^T\) is symmetric, prove!
Lectures 6-10
Vector Space
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All linear combination of vectors are in the space, called Vector Space
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Say \(P\) and \(L\) are two subspaces
- Is \(P\cup L\) a subspace? No, because some combination of vector \(A \in P\) and vector \(B \in L\) might not be in the subspace \(P\cup L\).
Column Space and Null Space
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Column Space C(\(A\)) is all possible linear combinations of column vectors in a matrix \(A\)
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Null Space N(\(A\)) is all the \(x\)’s for which \(Ax = 0\)
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Elimination process will change Column Space of \(A\), but not the Null space of \(A\)
- This is because in the elimination process we change \(A\), but not \(x\).
Basic and Free Variables
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The real difference between basic variables and free variables is that the free variables can be anything, and the basic variables are determined by solving the equations
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Basic variables are also called as Pivot variables. The columns with these variables are called Pivot columns
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Number of Pivot columns = Rank r of the matrix = Number of independent columns in the matrix
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Number of Free columns = n - r ( n = number of columns in matrix \(A))
Solution for Ax = b
Rank tells you everything about the number of solutions
- Full column rank matrix: Unique solution exists, if \(b\) lies in column space of \(A\)
- Full row rank matrix: Infinite number of solutions
- Full row rank/ full column rank: Unique solution always exists for all \(b\)
- No full column rank/ No full row rank: Infinitely many solution exists if \(b\) lies in column space of \(A\)
Basis
- Basis of a vector space is sequence of vectors \(v_1, v_2, v_3, …, v_N\) with the following properties
- They are independent
- They span a space
Four Subspaces
For a matrix \(A_{m*n}\)
- Column space C(\(A\)): combination of columns of \(A\), lies in \(R^n\)
- Null space N(\(A^T\)) : left null space of \(A\), lies in \(R^n\)
- Row space R(\(A\)): combination of columns of \(A^T\), lies in \(R^m\)
- Null space N(\(A\)): right null space of \(A\), lies in \(R^m\)
- Row operation on a matrix \(A\) changes the column space of \(A\), but preserves the row space of \(A\)
## Lectures 11-15
Vector Space of Matrices
- Consider vector space of matrices \(A_{N*N}\)
- Dimension of vector space of symmetric matrices = \((N^2 + N)/2\)
- Dimension of vector space of skew symmetric matrices = \((N^2 - N)/2\)
- Dimension of vector space of upper triangular matrices = \((N^2 + N)/2\)
- (Symmetric matrix \(\cap\) Upper triangular matrices) = Diagonal
matrices. Dimension of vector space = \((N^2 - N)/2\)
- (Symmetric matrix \(\cup\) Upper triangular matrices) = All \(N*N\) matrices. Dimension of vector space = \(N^2\)
Rank-1 Matrices
- They are special case
- They are building blocks for every other matrices
- Matrices with Rank-1 are separable matrices ( this is the
application in image processing for separating the filters)
Orthogonality
- Prove that two vectors \(X\) and \(Y\) are orthogonal to each other if
\(X^T Y = 0 = Y^T X\)
- Row space is perpendicular to Right null space , i.e., every vector
in row space is perpendicular to every other vector in right
null space
- Column space is perpendicular to left null space, i.e., every
vector in column space is perpendicular to every other vector
in left null space
Projection Matrix
- Projection of vector \(b\) onto vector \(a\) is defined as \(p = Pb\), where \(P = (a a^T)/(a^T a) \)
- Column space of projection matrix \(P\) is column space of \(a\). i.e., line through \(a\)
- Properties of Projection matrices
- Projection matrix in general
- We have \(A^T A \hat{x} = A^T B\)
- \(P = A \hat{x}\), where \(\hat{x} = (A^T A)^{-1} A^T B\)
and \(P = A (A^T A)^{-1} A^T \)
Least square problem
- Why projection ?
- When we have only few unknowns, but many equations to solve
\(Ax = b\), \(b\) might not be the column space of
\(A\)
- To solve for \(x\), we project \(b\) onto columns
space of \(A\), so that the new vector \(\bar{b}\) is closest
to \(b\)
- Eg. Measuring pulse of the heart, but repeatedly we take
measurements so that on an average we can estimate the pulse.
For more detailed treatment of least squares,
link
- The pictorial representation of solving least squares is shown below
- Least squares: \(\bar{x}\) minimizes \(b - Ax\) by solving \(A^T A \bar{x} = A^T b\)
- \(A^T A\) is invertible, if \(A\) has independent columns, prove
it