Recent content by squenshl

Right the basis for the kernel is the span of ##(1,1,1)##. Yes eigenvectors are linearly independent so they do span the range thanks!

So the basis for the range of ##T## are the other two eigenvectors.

Ok so the kernel of ##T## is ##(x,y,z)## such that ##T(x,y,z)=0## & this only occurs when we have ## (1,1,1)## so I guess that is the basis for the kernel right?

Okay so I found the eigenvalues to be ##\lambda = 0,-1,2## with corresponding eigenvectors ##v = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} ##. Not sure what to do next. Thanks!

Show that ##\{\mathbf{v}_1\}## is linearly independent. Simple enough let's consider $$c_1\mathbf{v}_1 = \mathbf{0}.$$ Our goal is to show that ##c_1 = 0##. By the definition of eigenvalues and eigenvectors we have ##A\mathbf{v}_1= \lambda_1\mathbf{v}_1##. Let's multiply the above equation and...

Thanks!

The solution is 3: It's just ##(g\circ f)'(6) = (-1,-2)\cdot (4,-3) = (-1\times 4)+((-2)\times (-3)) = -4+6 = 2## using the multi-variate chain rule and the dot product. Is this correct and if not how do I go about doing it? Thanks!

Okay then I’m lost how do we then justify whether to use the chain rule?

Okay (1) and (2) are done. So for (3), assuming ##t > 0##, ##f\circ \mathbf{r} = \ln{(e^{\sin{(t)}})}^2+\ln{(e^{\sin{(t)}})}^2 = \sin^2{(t)}+\cos^2{(t)} = 1## so the derivative is ##0##.

Yeah that certainly doesn't make sense! 1. Suppose ##\mathbf{0}\in T##, ##A(\mathbf{0})\in S## which is non-empty. 2. Suppose ##x_1,x_2\in T##. We then have that ##A(x_1+x_2) = A(x_1)+A(x_2)\in S##, i.e. ##x_1+x_2\in T##. 3. Suppose ##\mathbf{x}\in T##, with ##\lambda\in \mathbb{R}##...

Thanks! 1.Since ##\mathbf{0}\in T##, ##A(\mathbf{0})\in S## which is non-empty. 2. Suppose ##x_1,x_2\in T##. Then there exists vectors ##x_1,x_2\in S## such that we have ##A(x_1)## and ##A(x_2)##. We then have that ##x_1+x_2\in S## and ##A(x_1+x_2) = A(x_1)+A(x_2)##, i.e. ##x_1+x_2\in T##. 3...

The derivative of ##\mathbf{r}## at each point of ##(0,1)##?

1. We find the partial derivatives of ##f## with respect to ##x## and ##y## to get ##f_x = \frac{2\ln{(x)}}{x}## and ##f_y = \frac{2\ln{(y)}}{y}.## This makes the gradient vector $$\nabla{f} = \begin{bmatrix} f_x \\ f_y \end{bmatrix} = \begin{bmatrix} \frac{2\ln{(x)}}{x} \\ \frac{2\ln{(y)}}{y}...

1. Let's show the three conditions for a subspace are satisfied: Since ##\mathbf{0}\in \mathbb{R}^n##, ##A\times \mathbf{0} = \mathbf{0}\in S##. Suppose ##x_1, x_2\in \mathbb{R}^n##, then ##A(x_1+x_2) = A(x_1)+A(x_2)\in S##. Suppose ##x\in S## and ##\lambda\in \mathbb{R}##, then ##A(\lambda x) =...

There are sets of the form ##\left\{(x,y)\in \mathbb{R}^2: f(x,y) = \ln{\left(3+(x+y)^2\right)} = c\right\}## where ##c## is some fixed number ##> 1##. Let's see what happens for a few values of ##c##. Suppose ##c = 2##, then ##\ln{\left(3+(x+y)^2\right)} = 2 \Longleftrightarrow (x+y)^2 =...