Linear Algebra for Machine Learning and Data Science (14)

Determinants and Eigenvectors

Quiz

Q1

Let $T$ be a linear transformation in the plane represented by the following matrix:

$$ \left[\begin{array}{cc}1&0\\2&3\end{array}\right] $$

The rank of $T$ is:

1. 1
2. 0
3. 3
4. 2

Answer

4

At this point of the course, you have several ways of finding this information. Applying what you've seen in the lecture on Singularity and rank of linear transformations, it is necessary to understand how $T$ works in the vectors (0, 1) and (1,0).

$$ \left[\begin{array}{cc}1&0\\2&3\end{array}\right] \cdot \left[\begin{array}{c}0\\1\end{array}\right] = \left[\begin{array}{c}0\\3\end{array}\right] $$

and,

$$ \left[\begin{array}{cc}1&0\\2&3\end{array}\right] \cdot \left[\begin{array}{c}1\\0\end{array}\right] = \left[\begin{array}{c}1\\2\end{array}\right] $$

So, (0, 1) → (0, 3) and (1, 0) → (1, 2).

Note that vectors (0, 3) and (1, 2) form a parallelogram in the plane, therefore the rank of $T$ is 2

Q2

Consider the linear transformation $T$ that maps the vectors (1, 0) and (0, 1) in the following manner:

$$ T(0,1) = (2,5)\\ T(1,0)=(3,1) $$

The area of the parallelogram spanned by transforming the vectors (0, 1) and (1, 0) is:

Answer

13

The matrix associated with this linear transformation is given by

$$ \left[\begin{array}{cc}3&2\\1&5\end{array}\right] $$

The area of the parallelogram spanned by applying $T$ on the vectors (1,0) and (0,1) is then given by the determinant of such matrix:

$$ \text{det}\left(\left[\begin{array}{cc}3&2\\1&5\end{array}\right]\right) = 13 $$

You may watch again the lectures on Linear transformations as matrices to review how to obtain the matrix associated with a linear transformation and the lecture on Determinants as an area to review why the determinant represents the area of such a parallelogram.

Q3

Consider the following three matrices

$$ M_1 = \left[\begin{array}{cc}2&1\\3&1\end{array}\right]\\ M_2 = \left[\begin{array}{cc}3&5\\1&1\end{array}\right]\\ M_3 = \left[\begin{array}{cc}2&3\\4&5\end{array}\right] $$

The determinant of $M_1 \cdot M_2 \cdot M_3$ is equal to:

Answer

-4

As you've seen in the lecture Determinant of a product, the determinant of a product of matrices is the product of their determinant. Therefore:

$\det(M_1\cdot M_2 \cdot M_3) = \det(M_1)\cdot\det(M_2)\cdot\det(M_3)$

Since

$\det(M_1)=2-3=-1$

$\det(M_2)=3-5=-2$

$\det(M_3)=10-12=-2$

Then $\det(M_1\cdot M_2 \cdot M_3) = (-1)\cdot(-2)\cdot(-2) = -4$

Q4

Let $M$ and $N$ be two square matrices of the same size.

Check all true statements.

1. $\text{det}(M+N)=\text{det}(M)+\text{det}(N)$.
2. If $M$ is singular, then $M⋅N$ is singular for any matrix $N$.
3. If $M$ and $N$ are non-singular matrices, then so is $M⋅N$.
4. If $M⋅N$ is singular, then $M$ and $N$ are singular.

Answer

2

If $M$ is singular, then $\det(M) = 0$. So, for every matrix $N$ with the same size, $\det(M\cdot N) = \det(M)\cdot\det(N) = 0\cdot\det(N) = 0$. Therefore, $M\cdot N$ is singular

3

If $M$ and $N$ are non-singular, then $\det(M) \ne 0$ and $\det(N) \ne 0$. Sine $\det(M\cdot N) = \det(M)\det(N)$, then $\det(M\cdot N) \ne 0$ as well

Q5

Let $T$ be a linear transformation in the plane represented by the following matrix:

$$ \left[\begin{array}{cc}1&0\\2&3\end{array}\right] $$

The rank of $T$ is:

Answer

-0.5

As you've seen in the lecture Determinant of inverses, $\det(M^{-1}) = \det(M)^{-1} = {1\over \det(M)}$, so there is no need to compute the inverse of $M$! Computing the determinant for $M$:

$\det(M) = (0\cdot 2\cdot 0 + 0\cdot 1\cdot 1+1\cdot0\cdot2) - (1\cdot2\cdot1+0\cdot2\cdot0+0\cdot0\cdot1) = -2$

Therefore, $\det(M^{-1}) = {1\over -2} = -0.5$

All the information here is based on the Linear Algebra for Machine Learning and Data Science | Coursera from DeepLearning.AI