1.Let \(a\), \(b\) be two vectors in 2d real vector space \(\mathbb{R}_2\).
(a) If \(a = (1,2)\) and \(b = (1,4)\), are they linearly independent or not? If they are, can you write (4,4) as a linear combination of \(A\) and \(B\)?
(b) If \(a = (x,y)\) and \(b = (x,2y)\), what \(x\) and \(y\) value will make \(A\) and \(B\) linearly dependent?
2.Let \(c\), \(d\), and \(e\) be three vectors in 3d real vector space \(\mathbb{R}_3\).
(a) Let \(c=(1,2,3)\), \(d=(1,3,1)\), and \(e=(-1,-1,-5)\). Show that set \([c, d, e]\) cannot be a basis of \(\mathbb{R}_3\).
(b) Which vector in (a) you should change in order to obtain a basis of \(\mathbb{R}_3\)?
(c) Now, change the vector and prove it is a basis of \(\mathbb{R}_3\).
3.The polynomial \(P(x) = a_0 + a_1x + a_2x^2+ ... + a_nx^n\).
(a) Show that P(x) is a linear combination of some monomials.
(b) Can you find a basis for the polynomial vector space? If so, prove that they are indeed a basis for the space.
(c) What is the dimention of the polynomial vector space?
4.Orthonormal basis is a special kind of basis. It requires all basis vectors to be orthogonal to each other and has length of 1.
(a) Show that \(set[(1,0,0),(0,1,0),(0,0,1)]\) is a orthonormal basis for \(\mathbb{R}_3\)
(b) Find another orthonormal basis for \(\mathbb{R}_3\) (Try not to use the vectors on the axis)
Answer sheet
1.Let \(A\), \(B\) be two vectors in 2d real vector space \(\mathbb{R}_2\).
(a) If \(a = (1,2)\) and \(b = (1,4)\), are they linear independent or not? If they are, can you write (4,4) as a linear combination of \(A\) and \(B\)?
Just by looking at the vectors, you will see that they are not multiple of each other, so they are linear independent. To write \((4,4)\) in terms of \(a\) and \(b\), we can first write down \((1,0)\) and \((0,1)\) in terms of \(a\) and \(b\):
So, \((4,4)\) can be written as:
(b) If \(a = (x,y)\) and \(b = (x,2y)\), what \(x\) and \(y\) value will make \(A\) and \(B\) linear dependent?
Again, to let two vectors be linearly dependent in 2d, just make one to be multiple of the other. If we let \(x = 1\) and \(y = 0\), then \(a\) and \(b\) are the same vectors, therefore they are linearly dependent.
2.Let \(c\), \(d\), and \(e\) be three vectors in 3d real vector space \(\mathbb{R}_3\).
(a) Let \(c=(1,2,3)\), \(d=(1,3,1)\), and \(e=(-1,-1,-5)\). Show that set \([c, d, e]\) cannot be a basis of \(\mathbb{R}_3\).
To show all three of them are linearly independent is not easy, but to show they are linearly dependent is doable. Try to combine two of them and compare is a good strategy but is it not always guarantee. For our case, we can see that subtracting \(c\), \(d\) and adding \(c\),\(e\) will give you vectors that has the same x and y coordinate, so let’s start from there:
Well, we can see that \(d-c\) is the same as \(c+e\) so clearly they are not linearly independent. So they cannot form a basis of \(\mathbb{R}_3\).
(b) Which vector in (a) you should change in order to obtain a basis of \(\mathbb{R}_3\)?
Since none of the three vectors are multiple of any others directly, Change any of the three would work. But to show they are linearly independent(for now), you have to be able to write the basis vectors \((1,0,0),(0,1,0),(0,0,1)\) in terms of \(c\),\(d\),\(e\). in our case, we can change \(e\) to be \((-1,-1,-3)\) for convenient.
(c) Now, change the vector and prove it is a basis of \(\mathbb{R}_3\).
From part (a), we have:
Using our new \(e\) value, we have
So, to get \((0,0,1)\), we can subtract these two:
To get \((1,0,0)\), we can pick any of the vectors and subtract out the y and z components:
3.The polynomial \(P(x) = a_0 + a_1x + a_2x^2+ ... + a_nx^n\).
(a) Show that P(x) is a linear combination of some monomials.
Clearly, \(P(x)\) has the form of a linear combination:
If you treat the monomials \(x^n\) as \(\pmb{v}_n\).
(b) Can you find a basis for the polynomial vector space? If so, prove that they are indeed a basis for the space.
The basis will be:
(c) What is the dimension of the polynomial vector space?
The dimension of the polynomial vector space will be \(n+1\).
4.Orthonormal basis is a special kind of basis. It requires all basis vectors to be orthogonal to each other and has length of 1.
(a) Show that \(set[(1,0,0),(0,1,0),(0,0,1)]\) is a orthonormal basis for \(\mathbb{R}_3\)
\(set[(1,0,0),(0,1,0),(0,0,1)]\) is clearly a basis for \(\mathbb{R}_3\), so all we need to do is to check the length and orthogonality:
To check the orthogonality, we can use the dot product:
So they are orthogonal to each other. Therefore, \(set[(1,0,0),(0,1,0),(0,0,1)]\) is a orthonormal basis for \(\mathbb{R}_3\).
(b) Find another orthonormal basis for \(\mathbb{R}_3\) (Try not to use all the vectors on the axis)
The easiest way to do it is to have one vector stay the same and rotate the other two by a angle. In here, we use 45 degree for convenient:
First we check the lengths:
Then we check the orthogonality:
To have all three vectors not on the axis was not easy. To do that, we have to learn more techniques like Gram–Schmidt process or rotation matrix in 3d.