Tags: linear algebra, lecture-02, quiz-02
Let \(\vec x = (1, -2, 5, 0, 3, -1, 4)^T \in\mathbb{R}^7\). What is \(\vec x \cdot\hat{e}^{(3)}\)?
5.
The "right" way to do this problem is not to calculate the entire dot product, but to recognize that The dot product \(\vec x \cdot\hat{e}^{(3)}\) extracts the third component of \(\vec x\)(which is 5).
Tags: linear algebra, lecture-02, quiz-02
Let \(\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1)^T\) and \(\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(-1, 1)^T\) form an orthonormal basis \(\mathcal{U}\) for \(\mathbb{R}^2\).
Given \(\vec x = (4, 2)^T\), compute \([\vec x]_{\mathcal{U}}\). That is, compute the coordinates of \(\vec x\) in the new basis.
\([\vec x]_{\mathcal{U}} = (3\sqrt{2}, -\sqrt{2})^T\).
To find the coordinates of \(\vec x\) in the basis \(\mathcal{U}\), we compute the dot product of \(\vec x\) with each basis vector:
Therefore, the coordinates of \(\vec x\) in the basis \(\mathcal{U}\) are:
Tags: linear algebra, lecture-02, quiz-02
Let \(\hat{u}^{(1)} = \frac{1}{\sqrt{3}}(1, 1, 1)^T\), \(\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(1, -1, 0)^T\), and \(\hat{u}^{(3)} = \frac{1}{\sqrt{6}}(1, 1, -2)^T\) form an orthonormal basis \(\mathcal{U}\) for \(\mathbb{R}^3\).
Given \(\vec x = (3, 1, 2)^T\), compute \([\vec x]_{\mathcal{U}}\). That is, compute the coordinates of \(\vec x\) in the new basis.
\([\vec x]_{\mathcal{U}} = (2\sqrt{3}, \sqrt{2}, 0)^T\).
To find the coordinates of \(\vec x\) in the basis \(\mathcal{U}\), we compute the dot product of \(\vec x\) with each basis vector:
Therefore, the coordinates of \(\vec x\) in the basis \(\mathcal{U}\) are:
Tags: linear algebra, lecture-02, quiz-02
Let \(\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 0, 1)^T\), \(\hat{u}^{(2)} = (0, 1, 0)^T\), and \(\hat{u}^{(3)} = \frac{1}{\sqrt{2}}(1, 0, -1)^T\) form an orthonormal basis \(\mathcal{U}\) for \(\mathbb{R}^3\).
Given \(\vec x = (2, 3, 4)^T\), compute \([\vec x]_{\mathcal{U}}\). That is, compute the coordinates of \(\vec x\) in the new basis.
\([\vec x]_{\mathcal{U}} = (3\sqrt{2}, 3, -\sqrt{2})^T\).
To find the coordinates of \(\vec x\) in the basis \(\mathcal{U}\), we compute the dot product of \(\vec x\) with each basis vector:
Therefore, the coordinates of \(\vec x\) in the basis \(\mathcal{U}\) are:
Tags: linear algebra, lecture-02, quiz-02
Let \(\vec x = (2, -3)^T\) and let \(\mathcal{U} = \{\hat{u}^{(1)}, \hat{u}^{(2)}\}\) be an orthonormal basis for \(\mathbb{R}^2\). You are given the following information about the basis vectors:
Compute the coordinates of \(\vec x\) in the basis \(\mathcal{U}\), i.e., compute \([\vec x]_{\mathcal{U}}\).
\([\vec x]_{\mathcal{U}} = \left(-\frac{6}{5}, -\frac{17}{5}\right)^T\).
The coordinates of \(\vec x\) in basis \(\mathcal{U}\) are given by \([\vec x]_{\mathcal{U}} = (\vec x \cdot\hat{u}^{(1)}, \vec x \cdot\hat{u}^{(2)})^T\). We can compute each dot product using the given information:
Tags: linear algebra, lecture-02, quiz-02
Do the following three vectors form an orthonormal basis for \(\mathbb{R}^3\)? Justify your answer.
Nope, they don't. The problem is that the second and third vectors are not orthogonal. To be an orthonormal basis, all pairs of vectors must be orthogonal (dot product zero) and each vector must have unit length.
Tags: linear algebra, lecture-02, quiz-02
Do the following three vectors form an orthonormal basis for \(\mathbb{R}^3\)? Justify your answer.
Nope, they don't. While they are all orthogonal to each other, the second vector does not have unit length (its length is 2). To be an orthonormal basis, all vectors must have unit length.
Tags: linear algebra, lecture-02, quiz-02
Do the following three vectors form an orthonormal basis for \(\mathbb{R}^3\)? Justify your answer.
Yes.
You can check that all vectors have unit length, and that each pair of vectors is orthogonal (dot product zero).
Tags: lecture-02, linear transformations, quiz-02
Let \(\vec f(\vec x) = (2x_1 - x_2, \, x_1 + 3x_2)^T\). Is \(\vec f\) a linear transformation? Justify your answer.
Yes, \(\vec f\) is a linear transformation.
To verify linearity, we must show that for all vectors \(\vec x, \vec y\) and scalars \(\alpha, \beta\):
Let \(\vec x = (x_1, x_2)^T\) and \(\vec y = (y_1, y_2)^T\). Then:
Therefore, \(\vec f\) is linear.
Tags: lecture-02, linear transformations, quiz-02
Let \(\vec f(\vec x) = (x_1 x_2, \, x_2)^T\). Is \(\vec f\) a linear transformation? Justify your answer.
No, \(\vec f\) is not a linear transformation.
To show this, we find a counterexample. Consider \(\vec x = (1, 1)^T\) and \(\vec y = (1, 1)^T\), and let \(\alpha = \beta = 1\).
First, compute \(\vec f(\vec x + \vec y) = \vec f((2, 2)^T)\):
\(\vec f((2, 2)^T) = (2 \cdot 2, \, 2)^T = (4, 2)^T\) Now compute \(\vec f(\vec x) + \vec f(\vec y)\):
Since \((4, 2)^T \neq(2, 2)^T\), we have \(\vec f(\vec x + \vec y) \neq\vec f(\vec x) + \vec f(\vec y)\).
Therefore, \(\vec f\) is not linear. (The product \(x_1 x_2\) in the first component makes the function nonlinear.)
Tags: lecture-02, linear transformations, quiz-02
Let \(\vec f(\vec x) = (x_1^2, \, x_2)^T\). Is \(\vec f\) a linear transformation? Justify your answer.
No, \(\vec f\) is not a linear transformation.
A quick way to check: for any linear transformation, we must have \(\vec f(c\vec x) = c \vec f(\vec x)\) for any scalar \(c\).
Let us check the scaling property with \(\vec x = (1, 0)^T\) and \(c = 2\):
Since \((4, 0)^T \neq(2, 0)^T\), we have \(\vec f(2\vec x) \neq 2\vec f(\vec x)\).
Therefore, \(\vec f\) is not linear. (The squared term \(x_1^2\) makes the function nonlinear.)
Tags: lecture-02, linear transformations, quiz-02
Let \(\vec f(\vec x) = (-x_2, \, x_1)^T\). Is \(\vec f\) a linear transformation? Justify your answer.
Yes, \(\vec f\) is a linear transformation.
To verify linearity, we must show that for all vectors \(\vec x, \vec y\) and scalars \(\alpha, \beta\):
Let \(\vec x = (x_1, x_2)^T\) and \(\vec y = (y_1, y_2)^T\). Then:
Therefore, \(\vec f\) is linear.
By the way, this transformation does something interesting geometrically: it rotates vectors by \(90°\) counterclockwise.
Tags: lecture-02, linear transformations, quiz-02
Suppose \(\vec f\) is a linear transformation with:
Let \(\vec x = (2, 5)^T\). Compute \(\vec f(\vec x)\).
\(\vec f(\vec x) = (17, -1)^T\).
Since \(\vec f\) is linear, we can write:
Here, \(x_1 = 2\) and \(x_2 = 5\). Substituting the given values:
Tags: lecture-02, linear transformations, quiz-02
Suppose \(\vec f\) is a linear transformation with:
Let \(\vec x = (4, -2)^T\). Compute \(\vec f(\vec x)\).
\(\vec f(\vec x) = (-8, 10)^T\).
Since \(\vec f\) is linear, we can write:
Substituting the given values:
Tags: linear algebra, lecture-02, quiz-02
Recall that \(\vec x\) is a unit coordinate vector if the sum of the squares of its entries is 1. That is, if \(x_i\) is the \(i\)-th entry of \(\vec x\), then \(\vec x\) is a unit coordinate vector if \(\sum_i x_i^2 = 1\).
True or False: If \(\vec x\) is a unit coordinate vector when expressed in the standard basis, then \([\vec x]_{\mathcal{U}}\) is also a unit coordinate vector when expressed in an orthonormal basis \(\mathcal{U}\).
True.
The sum of squared entries of a vector is equal to the length of the vector squared. If we express \(\vec x\) in a different basis, we get different coordinates, but it doesn't change the length of the vector itself. So whatever those new coordinates are, their sum of squares must still equal 1.
Tags: linear algebra, lecture-02, quiz-02
True or False: If we express \(\vec x\) in an orthonormal basis \(\mathcal{U}\), the sum of the absolute values of the entries must stay the same. That is, \(\sum_i |x_i| = \sum_i |[\vec x]_{\mathcal{U},i}|\).
False.
The sum of absolute values (the \(\ell_1\) norm) is not preserved under a change of orthonormal basis. Only the sum of squares (the \(\ell_2\) norm squared) is preserved.
As a counterexample, consider \(\vec x = (1, 0)^T\) in \(\mathbb{R}^2\). The sum of absolute values is \(1\). Now consider the orthonormal basis \(\mathcal{U} = \{\frac{1}{\sqrt{2}}(1, 1)^T, \frac{1}{\sqrt{2}}(-1, 1)^T\}\). Then:
The sum of absolute values is now \(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\neq 1\).
Tags: linear algebra, lecture-02, quiz-02
Let \(\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1)^T\) and \(\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(1, -1)^T\) form an orthonormal basis \(\mathcal{U}\) for \(\mathbb{R}^2\).
Suppose \([\vec x]_{\mathcal{U}} = (\sqrt{2}, 3\sqrt{2})^T\). That is, the coordinates of \(\vec x\) in the basis \(\mathcal{U}\) are \((\sqrt{2}, 3\sqrt{2})^T\).
What is \(\vec x\) in the standard basis?
\(\vec x = (4, -2)^T\).
To convert from basis \(\mathcal{U}\) to the standard basis, we compute the linear combination of basis vectors: