Sean Fitzpatrick |
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University of Lethbridge |
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Show that if \(A^3 = 4I_n\text{,}\) then \(A\) is invertible.
Show that if \(A^2-3A+2I_n=0\text{,}\) then \(A\) is invertible.
Simplify the expression \(B(AB)^{-1}(AB)^2B^{-1}\text{.}\)
Find the inverse of \(A = \bbm 1\amp 4\\2\amp 5\ebm\text{.}\)
Let \(A\) be an \(n\times n\) matrix. The following statements are all equivalent to the statement “\(A\) is invertible”:
The rank of \(A\) is \(n\text{.}\)
The RREF of \(A\) is equal to \(I_n\text{.}\)
The system \(A\vec{x}=\vec{b}\) is consistent for every \(n\times 1\) vector \(\vec{b}\text{.}\)
The only solution to the system \(A\vec{x}=\vec{0}\) is \(\vec{x}=\vec{0}\text{.}\)
There is a matrix \(B\) such that \(AB=I_n\text{.}\)
There is a matrix \(C\) such that \(CA=I_n\text{.}\)
Let \(A\) and \(B\) be invertible matrices.
The inverse of \(A\) is unique.
\(A^{-1}\) is also invertible, and \((A^{-1})^{-1}=A\text{.}\)
\(AB\) is invertible, and \((AB)^{-1}=B^{-1}A^{-1}\text{.}\)
For any nonzero scalar \(k\text{,}\) \(kA\) is invertible, and \((kA)^{-1}=\frac1k A^{-1}\text{.}\)
For any \(n\times 1\) vector \(\vec{b}\text{,}\) the unique solution to \(A\vec{x}=\vec{b}\) is \(\vec{x}=A^{-1}\vec{b}\text{.}\)
Recall: if \(A\) is an \(m\times n\) matrix, and \(\vec{x}\) is an \(n\times 1\) vector, then \(\vec{y}=A\vec{x}\) is an \(m\times 1\) vector.
Let \(\R^n\) (sometimes, \(\R^{n,1}\)) denote the set of all \(n\times 1\) column vectors.
Then any \(m\times n\) matrix \(A\) can be used to define a function
What properties does this function have?
Let \(A = \bbm 3\amp -2\\5\amp 4\ebm\text{,}\) and let \(T(\vec{x})=A\vec{x}\text{.}\)
Compute the value of \(T\) on:
Repeat the above for \(A = \bbm 1\amp -2\\-2\amp 4\ebm\text{.}\)
Let \(A\) be an \(m\times n\) matrix, and define \(T(\vec{x})=A\vec{x}\text{.}\) Then:
\(T\) is a function from \(\R^n\) to \(\R^m\)
\(T(\vec{x}+\vec{y})=T(\vec{x})+T(\vec{y})\) for all \(\vec{x},\vec{y}\)
\(T(c\vec{x})=cT(\vec{x})\) for all vectors \(\vec{x}\) and scalars \(c\)
Note: any function between vector spaces (e.g. \(\R^n,\R^m\)) with these properties is called a linear transformation.
It turns out any linear transformation can be expressed as a matrix transformation.
Show that for any linear transformation, \(T(\vec{0})=\vec{0}\text{.}\)
Given \(T\left(\bbm 1\\0\ebm\right) = \bbm 3\\-2\ebm\) and \(T\left(\bbm 0\\1\ebm\right) = \bbm -4\\5\ebm\text{,}\) find \(T\left(\bbm 2\\3\ebm\right)\)
Given \(T(\vec{a})=\bbm 1\\0\\-3\ebm\text{,}\) \(T(\vec{b})=\bbm 0\\-2\\5\ebm\) and \(T(\vec{c})=\bbm 1\\1\\1\ebm\text{,}\) find \(T(2\vec{a}-3\vec{b}+5\vec{c})\text{.}\)
If \(T(\vec{x})=A\vec{x}\) for \(A = \bbm 2\amp -3\\-1\amp -2\\5\amp 4\ebm\text{,}\) what are the domain and codomain of \(T\text{?}\)
For \(T\) as above, compute \(T(\hat{\imath})\) and \(T(\hat{\jmath})\)
What if \(A = \bbm 2\amp -1\amp 5\\4\amp -2\amp 3\ebm\text{?}\)
If \(T\left(\bbm x\\y\\z\ebm\right) = \bbm 3x-2y\\-x+4y+5z\\7x-2y-6z\ebm\text{,}\) for what matrix is \(T(\vec{x})=A\vec{x}\text{?}\)
When \(A\) is \(2\times 2\) we can visualize everything in terms of geometric vectors in the plane.
We can use matrices to describe transformations, like stretches, rotations, and reflections.
(But not translations.)
Describe the effect of the transformation with matrix \(A = \bbm 2\amp -1\\1\amp 3\ebm\) in terms of what it does to the unit square (\(0\leq x,y\leq 1\))
Stretches: \(\bbm k\amp 0\\0\amp 1\ebm, \bbm 1\amp 0\\0\amp k\ebm, \bbm k\amp 0\\0\amp k\ebm = kI_2\text{.}\) (This is just scalar multiplication.)
Reflections: \(\bbm -1\amp 0\\0\amp 1\ebm, \bbm 1\amp 0\\0\amp -1\ebm, \bbm -1\amp 0\\0\amp -1\ebm, \bbm 0\amp 1\\1\amp 0\ebm\)
Rotations: \(\bbm \cos(\theta)\amp -\sin(\theta)\\ \sin(\theta)\amp \cos(\theta)\ebm\)
Shears: \(\bbm 1\amp k\\0\amp 1\ebm, \bbm 1\amp 0\\k\amp 1\ebm\)
Determine the matrix transformation that:
Stretches horizontally by a factor of 2, rotates by \(90^\circ\text{,}\) and then reflects across the \(x\) axis.
Reflects across the line \(y=x\text{,}\) stretches vertically be a factor of 3, then reflects across the \(y\) axis.