Assume that T is a linear transformation. Find the standard matrix of T.

Burhan Hopper
2020-12-28
Answered

Assume that T is a linear transformation. Find the standard matrix of T.

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gotovub

Answered 2020-12-29
Author has **98** answers

Let

1.Calculate

2.Find the coordinate vectors

3.Write the matrix with columns as the column vectors calculated in Step 2:

Clearly, here

Now we calculate

Now we create the matrix M with columns as the column as the column vectors of

Therefore

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Goes through $(4,5)$ and parallel to the x-axis

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Goes through $(-1,-2)$ and is parallel to the line $2x+3y=8$

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The Determinant, Tensors, and Orientation

I am a bit confused about orientation and tensors as exemplified by the determinant.

If we have an inner product or a metric and transform a vector from one coordinate system to another the magnitude of that vector is unchanged as the coordinate representation of the metric also changes. Therefore it is natural to think of a vector as a tensor and it's length doesn't change under orientation reversing transformations.

In contrast, we can consider the determinant as an alternating tensor from the exterior algebra, for instance in ${\mathbb{R}}^{3}$,

$det({v}_{1},{v}_{2},{v}_{3})={e}^{1}\wedge {e}^{2}\wedge {e}^{3}({v}_{1},{v}_{2},{v}_{3})$

But if we perform a reflection, A then we get,

$det(A{v}_{1},A{v}_{2},A{v}_{3})=det(A)det({v}_{1},{v}_{2},{v}_{3})=-det({v}_{1},{v}_{2},{v}_{3})$

Does that mean that the 3-covector ${e}^{1}\wedge {e}^{2}\wedge {e}^{3}$ when contracted with 3 vectors is not invariant, which would seem to violate the idea of contracting tensors to a scalar produces an invariant under transformations.

Or do we instead by transforming our 3-covector and the input vectors get an invariant oriented volume measure?

${e}^{\prime 1}\wedge {e}^{\prime 2}\wedge {e}^{\prime 3}(A{v}_{1},A{v}_{2},A{v}_{3})={e}^{1}\wedge {e}^{2}\wedge {e}^{3}({v}_{1},{v}_{2},{v}_{3})?$

That would seem to mean that the determinant includes an arbitrary sign convention and that the definition of the determinant is different in left and right handed oriented coordinate systems.

I am a bit confused about orientation and tensors as exemplified by the determinant.

If we have an inner product or a metric and transform a vector from one coordinate system to another the magnitude of that vector is unchanged as the coordinate representation of the metric also changes. Therefore it is natural to think of a vector as a tensor and it's length doesn't change under orientation reversing transformations.

In contrast, we can consider the determinant as an alternating tensor from the exterior algebra, for instance in ${\mathbb{R}}^{3}$,

$det({v}_{1},{v}_{2},{v}_{3})={e}^{1}\wedge {e}^{2}\wedge {e}^{3}({v}_{1},{v}_{2},{v}_{3})$

But if we perform a reflection, A then we get,

$det(A{v}_{1},A{v}_{2},A{v}_{3})=det(A)det({v}_{1},{v}_{2},{v}_{3})=-det({v}_{1},{v}_{2},{v}_{3})$

Does that mean that the 3-covector ${e}^{1}\wedge {e}^{2}\wedge {e}^{3}$ when contracted with 3 vectors is not invariant, which would seem to violate the idea of contracting tensors to a scalar produces an invariant under transformations.

Or do we instead by transforming our 3-covector and the input vectors get an invariant oriented volume measure?

${e}^{\prime 1}\wedge {e}^{\prime 2}\wedge {e}^{\prime 3}(A{v}_{1},A{v}_{2},A{v}_{3})={e}^{1}\wedge {e}^{2}\wedge {e}^{3}({v}_{1},{v}_{2},{v}_{3})?$

That would seem to mean that the determinant includes an arbitrary sign convention and that the definition of the determinant is different in left and right handed oriented coordinate systems.

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$(2,-1),(-\frac{1}{3},\frac{4}{5})$

$(2,-1),(-\frac{1}{3},\frac{4}{5})$

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Spherical: $(8,\frac{\pi}{3},\frac{\pi}{6})$
Change to cylindrical.

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A line L through the origin in