Change of basis基底変換きていへんかん and similarity相似そうじ - Basic Exercises

1Exercise plan

A change of basis基底変換きていへんかん changes the coordinate representation座標表示ざひょうひょうじ, not the vectorベクトル itself or the linear map線型写像せんけいしゃぞう itself. Distinguish old-basis coordinates, new-basis coordinates, and representation matrices表現行列ひょうげんぎょうれつ.

Keep the convention fixed throughout: in these problems, $P$P sends new-basis coordinates to old-basis coordinates, so $[x{]}_{\mathrm{o}\mathrm{l}\mathrm{d}}=P[x{]}_{\mathrm{n}\mathrm{e}\mathrm{w}}$[x]_{\mathrm{old}}=P[x]_{\mathrm{new}}. This is why a representation matrix changes as $P^{-1}{A}_{\mathrm{o}\mathrm{l}\mathrm{d}}P$P^{-1}A_{\mathrm{old}}P. Reversing the convention reverses the formula, so the order cannot be treated as cosmetic.

For rectangular maps, use separate basis changes on the domain and codomain and expect $Q^{-1}AP$Q^{-1}AP. For every proposed change-of-basis matrix, first check invertibility; if $P$P is singular, the columns do not form a basis and the operation collapses information instead of merely relabeling coordinates.

Determinants and eigenvalues in this exercise are previews for later exercise sets. The main change-of-basis computations check invertibility by column independence or by an explicit inverse matrix.

2Check problem: change-of-coordinates matrix座標変換行列ざひょうへんかんぎょうれつ

Let the standard basis標準基底ひょうじゅんきてい of $R^2$\mathbb R^2 be the old basis, and let the new basis基底きてい be

Find the new-basis coordinates $[x{]}_{\mathrm{n}\mathrm{e}\mathrm{w}}$[x]_{\mathrm{new}} of

2.1Answer

Put the new basis vectors in the columns:

This matrix sends new coordinates to old coordinates:

Solve

Then

so $s=2$s=2 and $t=1$t=1. Therefore

The vector is unchanged; only its coordinate description changes.

3Check problem: similarity transformation相似変換そうじへんかん

Let

Find $A_{\mathrm{n}\mathrm{e}\mathrm{w}}=P^{-1}{A}_{\mathrm{o}\mathrm{l}\mathrm{d}}P$A_{\mathrm{new}}=P^{-1}A_{\mathrm{old}}P.

3.1Answer

The determinant check is a preview. Equivalently, the two columns are linearly independent, and the displayed inverse can be verified by multiplication.

Since $\det P=-2\ne 0$\det P=-2\ne0,

Also

Thus

3.2Explanation

The right-side $P$P converts a new-coordinate input to old coordinates. $A_{\mathrm{o}\mathrm{l}\mathrm{d}}$A_{\mathrm{old}} applies the map in old coordinates. The left-side $P^{-1}$P^{-1} converts the result back to new coordinates.

4Check problem: changing bases in the domain定義域ていぎいき and codomain終域しゅういき

For $T:R^2\to R^3$T:\mathbb R^2\to\mathbb R^3, let

Find $A_{\mathrm{n}\mathrm{e}\mathrm{w}}=Q^{-1}{A}_{\mathrm{o}\mathrm{l}\mathrm{d}}P$A_{\mathrm{new}}=Q^{-1}A_{\mathrm{old}}P.

4.1Answer

Since $Q=I_3$Q=I_3,

4.2Explanation

For a rectangular matrix長方行列ちょうほうぎょうれつ, the form is $Q^{-1}AP$Q^{-1}AP rather than similarity, because domain and codomain have separate bases.

5Check problem: quantities preserved by similarity

For

confirm that traceトレース, determinant行列式ぎょうれつしき, and eigenvalues固有値こゆうち agree.

Preview note: determinants and eigenvalues are developed later. Here this is an advanced check using the $2$2 by $2$2 determinant formula and the fact that the eigenvalues of an upper triangular matrix are its diagonal entries.

5.1Answer

Both are upper triangular, so their eigenvalues are the diagonal entries:

The eigenvalues agree up to order.

5.2Explanation

Similarity is an alternative representation of the same linear map. Entries may change, but basis-independent quantities are preserved.

6Check problem: checking invertibility可逆性かぎゃくせい

Can

be used as a change-of-coordinates matrix?

6.1Answer

No. The second column is twice the first:

Thus the columns are linearly dependent線型従属せんけいじゅうぞく. Also

so $P^{-1}$P^{-1} does not exist.

6.2Explanation

A non-invertible matrix collapses information. A change of basis must only relabel coordinates, so its matrix must be invertible.

For a change of basis, the columns of $P$P are the new basis vectors written in the old coordinates, so they must be linearly independent線型独立せんけいどくりつ. The determinant calculation confirms the same obstruction: using the inverse formula would require division by $\det P$\det P, and here that value is $0$0. Thus this matrix is not a coordinate-change matrix.

The determinant explanation is a preview. The main reason is already visible from the columns: the second column is twice the first, so the two columns cannot form a basis.

7Supplementary problems: failure examples and order

7.1Problem 6

Explain from invertibility why

are not a basis of $R^2$\mathbb R^2.

7.2Answer example

Putting them in columns gives

which is not invertible. Coordinates would not recover vectors uniquely, so the vectors are not a basis.

7.3Problem 7

For

compute $P^{-1}AP$P^{-1}AP and $PA{P}^{-1}$PAP^{-1}.

7.4Answer example

Then

They are not the same, so the order cannot generally be swapped.

Problem 6 checks that using a non-basis breaks coordinate change: the matrix built from $b_1,b_2$b_1,b_2 collapses the plane and cannot recover vectors uniquely. Problem 7 checks the order in a similarity transformation相似変換そうじへんかん. The expressions $P^{-1}AP$P^{-1}AP and $PA{P}^{-1}$PAP^{-1} look similar, but they describe opposite coordinate conventions and generally give different matrices.

8Related lectures

9Summary

• In a change of basis基底変換きていへんかん, the coordinate representation changes, not the vector itself.
• A similarity transformation相似変換そうじへんかん $P^{-1}AP$P^{-1}AP represents the same linear map in another basis.
• Changing bases in the domain and codomain separately gives $Q^{-1}AP$Q^{-1}AP.
• Whenever $P^{-1}$P^{-1} is used, confirm that $P$P is invertible可逆かぎゃく.