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Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad A\hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of reversing aflipping the sign on basis vectorvectors: upside down map

Instead of a map where North is upthe y-axis measures how far north and the x-axis measures how far east, you could just as easily have a map where South is upthe y-axis measures how far south and the x-axis measures how far west.

Up and down would still be aligned with the magnetic axis. In some mathematical sense, it's entirely equivalent. The only difference is perhaps psychological.

alt text

Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad A\hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of reversing a basis vector: upside down map

Instead of a map where North is up, you could just as easily have a map where South is up.

Up and down would still be aligned with the magnetic axis. In some mathematical sense, it's entirely equivalent. The only difference is perhaps psychological.

alt text

Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad A\hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of flipping the sign on basis vectors: upside down map

Instead of a map where the y-axis measures how far north and the x-axis measures how far east, you could just as easily have a map where the y-axis measures how far south and the x-axis measures how far west.

Up and down would still be aligned with the magnetic axis.

alt text

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Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad \hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad A\hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of reversing a basis vector: upside down map

Instead of a map where North is up, you could just as easily have a map where South is up.

Up and down would still be aligned with the magnetic axis. In some mathematical sense, it's entirely equivalent. The only difference is perhaps psychological.

alt text

Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad \hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of reversing a basis vector: upside down map

Instead of a map where North is up, you could just as easily have a map where South is up.

Up and down would still be aligned with the magnetic axis. In some mathematical sense, it's entirely equivalent. The only difference is perhaps psychological.

alt text

Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad A\hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of reversing a basis vector: upside down map

Instead of a map where North is up, you could just as easily have a map where South is up.

Up and down would still be aligned with the magnetic axis. In some mathematical sense, it's entirely equivalent. The only difference is perhaps psychological.

alt text

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source | link

Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad \hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of reversing a basis vector: upside down map

Instead of a map where North is up, you could just as easily have a map where South is up.

Up and down would still be aligned with the magnetic axis. In some mathematical sense, it's entirely equivalent. The only difference is perhaps psychological.

alt text

Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad \hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

Something that various people are pointing out is that the vectors $(1,1)$, $(2,2)$, or $(-1,-1)$ all represent the same line. When you find an eigenvector, what's uniquely determined is the line, not the actual vector.

An eigenvector for a matrix (linear transformation) $A$ is defined as any vector $\mathbf{v} \neq \mathbf{0}$ which satisfies:

$$A \mathbf{v} = \lambda \mathbf{v}$$

If $\mathbf{v}$ is an eigenvector, any scalar multiple $\hat{\mathbf{v}} = \alpha \mathbf{v}$ will also work ($\alpha \neq 0$):

\begin{align*} A \mathbf{v} = \lambda \mathbf{v}\quad & \Leftrightarrow \quad \alpha A \mathbf{v} = \alpha \lambda \mathbf{v} \\ & \Leftrightarrow \quad \hat{\mathbf{v}} = \lambda \hat{\mathbf{v}} \end{align*}

Eg. Choose $\alpha = -1$. If $\mathbf{v}$ is an eigenvector, so is $-\mathbf{v}$.

Let's say your PCA algorithm guarantees you that $\|\mathbf{v}\| = 1$. You still have two possibilities because if you take the intersection of a line through the origin and the unit circle, you get two points.

enter image description here

In this example, whether you have $\mathbf{v} = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\mathbf{v} = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, it really doesn't matter.

A fun example of reversing a basis vector: upside down map

Instead of a map where North is up, you could just as easily have a map where South is up.

Up and down would still be aligned with the magnetic axis. In some mathematical sense, it's entirely equivalent. The only difference is perhaps psychological.

alt text

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