# Linear Algebra - Gram-Schmidt Process

• joe_cool2
In summary, the conversation discusses using the Gram-Schmidt process to transform a basis {u1,u2} into an orthonormal basis in R2 with the Euclidean inner product. The process involves finding the norm of u1 and using it to normalize v2. However, there was an error in the calculation of the norm of v2, resulting in an incorrect answer. To verify the correctness of the solution, one can check if the dot products of the generated set of vectors are all zero and also confirm that the final vectors have a norm of one.
joe_cool2

## Homework Statement

Let R2 have the Euclidean inner product and use the Gram-Schmidt process to transform the basis {u1,u2} into an orthonormal basis.

u1 = (1,-3)
u2 = (2,2)

## Homework Equations

Gram-Schmidt process:

$$\\v_1 = u_1 \\v_2= u_2 - \frac{\left ( \left \langle u_2, v_1\right \rangle \right )}{\left \| v_1 \right \| ^ 2}v_1$$

## The Attempt at a Solution

We'll go ahead and find the norm of v first. Then we'll evaluate the second Gram-Schmidt equation.

$$\\\left \| v_1 \right \| = \sqrt{10} \\\\ \left \| v_1 \right \|v_1 = u_1 = (1,-3) \\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \left \langle \begin{bmatrix} 2\\2 \end{bmatrix} , \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \rangle \right )}{\left \| \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \| ^ 2} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix}\\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \frac{-4}{\sqrt{10}} \right )}{1} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \\\\ \begin{bmatrix} 2\\2 \end{bmatrix} - \begin{bmatrix} \frac{-4}{10}\\\\\frac{12}{10} \end{bmatrix} = \begin{bmatrix} \frac{24}{10}\\\\\frac{8}{10} \end{bmatrix} = \begin{bmatrix} \frac{12}{5}\\\\\frac{4}{5} \end{bmatrix}$$

I've not been able to get another answer for this step. Even checked in Matlab.

Now, we'll normalize the vector like the problem wants us to:
$$\\\left \| v_2 \right \| = \sqrt{8} \\ v_2 = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{12}{10\sqrt{2}}\\\\ \frac{4}{10\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6}{5\sqrt{2}}\\\\ \frac{2}{5\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6\sqrt{2}}{10}\\\\ \frac{2\sqrt{2}}{10} \end{bmatrix} = \begin{bmatrix} \frac{3\sqrt{2}}{5}\\\\ \frac{\sqrt{2}}{5} \end{bmatrix}$$

This is an odd problem, so I know that the solution should be $v_2 = (\frac{3}{\sqrt{10}} , \frac{1}{\sqrt{10}})$. There must be something wrong when I'm finding $||v_2||v_2$ . Is there anything fundamentally wrong with my process?

Last edited:
joe_cool2 said:

## Homework Statement

Let R2 have the Euclidean inner product and use the Gram-Schmidt process to transform the basis {u1,u2} into an orthonormal basis.

u1 = (1,-3)
u2 = (2,2)

## Homework Equations

Gram-Schmidt process:

$$\\v_1 = u_1 \\v_2= u_2 - \frac{\left ( \left \langle u_2, v_1\right \rangle \right )}{\left \| v_1 \right \| ^ 2}v_1$$

## The Attempt at a Solution

We'll go ahead and find the norm of v first. Then we'll evaluate the second Gram-Schmidt equation.

$$\\\left \| v_1 \right \| = \sqrt{10} \\\\ \left \| v_1 \right \|v_1 = u_1 = (1,-3) \\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \left \langle \begin{bmatrix} 2\\2 \end{bmatrix} , \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \rangle \right )}{\left \| \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \| ^ 2} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix}\\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \frac{-4}{\sqrt{10}} \right )}{1} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \\\\ \begin{bmatrix} 2\\2 \end{bmatrix} - \begin{bmatrix} \frac{-4}{10}\\\\\frac{12}{10} \end{bmatrix} = \begin{bmatrix} \frac{24}{10}\\\\\frac{8}{10} \end{bmatrix} = \begin{bmatrix} \frac{12}{5}\\\\\frac{4}{5} \end{bmatrix}$$

I've not been able to get another answer for this step. Even checked in Matlab.
If you get rid of a common factor of 4/5, you have ##\vec{v}_2 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}##.

Now, we'll normalize the vector like the problem wants us to:
$$\\\left \| v_2 \right \| = \sqrt{8} \\ v_2 = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{12}{10\sqrt{2}}\\\\ \frac{4}{10\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6}{5\sqrt{2}}\\\\ \frac{2}{5\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6\sqrt{2}}{10}\\\\ \frac{2\sqrt{2}}{10} \end{bmatrix} = \begin{bmatrix} \frac{3\sqrt{2}}{5}\\\\ \frac{\sqrt{2}}{5} \end{bmatrix}$$

This is an odd problem, so I know that the solution should be $v_2 = (\frac{3}{\sqrt{10}} , \frac{1}{\sqrt{10}})$. There must be something wrong when I'm finding $||v_2||v_2$ . Is there anything fundamentally wrong with my process?

Ok. So I gather that my answer is not technically wrong? How can I verify that my original answer is in fact an orthonormal basis spanning the same subspace?

I guess if I make computational errors, it will be unlikely that all the dot products in some generated set of vectors is zero. Right?

Last edited:

Ah, I see that I accidentally used the norm of u2 instead of v2 to normalize v2. The book confused me by switching up u's and v's.

Checking to see if the norm of the final vectors is one would also be a useful trick.

## 1. What is the Gram-Schmidt process?

The Gram-Schmidt process is a method used in linear algebra to orthogonalize a set of vectors. This means converting a set of vectors that may not be perpendicular to each other into a set of vectors that are perpendicular.

## 2. Why is the Gram-Schmidt process important?

The Gram-Schmidt process is important because it allows us to simplify calculations involving linear combinations of vectors. It also helps us find a basis for a vector space, which is useful in solving a variety of problems in mathematics and engineering.

## 3. How does the Gram-Schmidt process work?

The Gram-Schmidt process involves three steps. First, we take the first vector in the set and make it the first basis vector. Then, we take the second vector and subtract its projection onto the first vector, creating a new vector that is orthogonal to the first. This new vector becomes the second basis vector. We repeat this process for the remaining vectors in the set, creating a set of orthogonal basis vectors.

## 4. What is the significance of orthogonal basis vectors?

Orthogonal basis vectors are significant because they make calculations involving linear combinations of vectors simpler. When a set of vectors is orthogonal, the coefficients in a linear combination can be easily determined by taking the dot product of the vector with each basis vector. Additionally, orthogonal basis vectors are useful in solving systems of linear equations and finding eigenvalues and eigenvectors.

## 5. Can the Gram-Schmidt process be applied to any set of vectors?

Yes, the Gram-Schmidt process can be applied to any set of vectors in any vector space. However, it is most commonly used in Euclidean spaces, where the concept of orthogonality is well-defined. In abstract vector spaces, the Gram-Schmidt process may involve more complex calculations.

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